Excellent Arrays Codeforces Solution

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Excellent Arrays Codeforces Solution

Let’s call an integer array a1,a2,…,ana1,a2,…,an good if ai≠iai≠i for each ii. Let F(a)F(a) be the number of pairs (i,j)(i,j) (1≤i<j≤n1≤i<j≤n) such that ai+aj=i+jai+aj=i+j.

Let’s say that an array a1,a2,…,ana1,a2,…,an is excellent if:

  • aa is good;
  • l≤ai≤rl≤ai≤r for each ii;
  • F(a)F(a) is the maximum possible among all good arrays of size nn.

Given nn, ll and rr, calculate the number of excellent arrays modulo 109+7109+7.Input

The first line contains a single integer tt (1≤t≤10001≤t≤1000) — the number of test cases.

The first and only line of each test case contains three integers nn, ll, and rr (2≤n≤2⋅1052≤n≤2⋅105; −109≤l≤1−109≤l≤1; n≤r≤109n≤r≤109).

It’s guaranteed that the sum of nn doesn’t exceed 2⋅1052⋅105.Output

For each test case, print the number of excellent arrays modulo 109+7109+7.

Example

input

4
3 0 3
4 -3 5
42 -33 55
69 -42 146

output

4
10
143922563
698570404

Note

In the first test case, it can be proven that the maximum F(a)F(a) among all good arrays aa is equal to 22. The excellent arrays are:

  1. [2,1,2][2,1,2];
  2. [0,3,2][0,3,2];
  3. [2,3,2][2,3,2];
  4. [3,0,1][3,0,1].

Solution

#include <bits/stdc++.h>
using namespace std;

#define debug(...) do {\
	fprintf(stderr, "%s - %d : (%s) = ", __PRETTY_FUNCTION__, __LINE__, #__VA_ARGS__);\
    _DO(__VA_ARGS__);\
}while(0) 
template<typename I> void _DO(I&&x) {cerr << x << '\n';}
template<typename I, typename ...T> void _DO(I&&x,T&&...tail) {cerr << x << ", "; _DO(tail...);}
#define IOS
#else
#define debug(...)
#define IOS ios_base::sync_with_stdio(0);cin.tie(0)
#endif
#define FOR(i, n) for(int i = 0; i < n; i++)
#define FOR1(i, n) for(int i = 1; i <= n; i++)
#define pb push_back
#define mp make_pair
#define rsz resize
#define X first
#define Y second
#define SZ(x) (ll)x.size()
#define ALL(x) (x).begin(),(x).end()
#define SORT(x) sort(ALL(x))
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vll;
typedef vector<vll> vvll;
const int NF = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MOD = 1e9 + 7;
const ld PI = 3.14159265358979323846264338327950288;

ll fact[200005], invfact[200005];
ll mypow(ll a, ll x){
    if (x == 0) return 1;
    if (x == 1) return a;
    ll tmp = mypow(a, x >> 1);
    tmp = tmp * tmp % MOD;
    if (x & 1) tmp *= a;
    return tmp % MOD;
}
ll inv(ll a){
    return mypow(a, MOD - 2);
}
ll C(ll n, ll k){
    return fact[n] * invfact[k] % MOD * invfact[n - k] % MOD;
}
ll n, l, r;
ll sub(ll p, ll m, ll x){
    if (p < l + x - 1) return 0;
    if (m < n - (r - x + 1) + 1) return 0;
    p -= max(l + x - 1LL, 0LL);
    m -= max((ll)(n - (r - x + 1) + 1), 0LL);
    return C(m + p, p);
}
int main() {
    IOS;
    int t;
    cin >> t;
    fact[0] = 1;
    invfact[0] = 1;
    for (ll i = 1; i < 200005; i++){
        fact[i] = fact[i - 1] * i % MOD;
        invfact[i] = inv(fact[i]);
    }
    while (t--){
        cin >> n >> l >> r;
        ll ans = min(r - n, 1LL - l) * ((n & 1) ? C(n, n / 2) + C(n, n / 2 + 1) : C(n, n / 2));
        ans %= MOD;

        for (ll x = min(r - n, 1LL - l) + 1; x <= min(r - 1LL, n - l); x++){
            if (l + x - 1 >= r - x + 1) break;

            if (n & 1){
                ans += sub(n / 2, n / 2 + 1, x);
                ans %= MOD;
                ans += sub(n / 2 + 1, n / 2, x);
                ans %= MOD;
            }
            else {
                ans += sub(n / 2, n / 2, x);
                ans %= MOD;
            }
        }
        cout << ans << '\n';
    }
    return 0;
}

Educational Codeforces Round 111 (Rated for Div. 2)

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