Problem: Devu wants to decorate his garden with flowers. He has purchased n boxes, where the i-th box contains fi flowers. All flowers in a single box are of the same color (hence they are indistinguishable). Also, no two boxes have flowers of the same color.

Now Devu wants to select exactly s flowers from the boxes to decorate his garden. Devu would like to know, in how many different ways can he select the flowers from each box? Since this number may be very large, he asks you to find the number modulo (109 + 7). 

Devu considers two ways different if there is at least one box from which different number of flowers are selected in these two ways.

Input

The first line of input contains two space-separated integers n and s (1 ≤ n ≤ 20, 0 ≤ s ≤ 1014).

The second line contains n space-separated integers f1, f2, ... fn (0 ≤ fi ≤ 1012).

Output

Output a single integer — the number of ways in which Devu can select the flowers modulo (109 + 7).

Examples

Input

2 3
1 3


Output

2


Input

2 4
2 2


Output

1


Input

3 5
1 3 2


Output

3

Note

Sample 1. There are two ways of selecting 3 flowers: {1, 2} and {0, 3}.

Sample 2. There is only one way of selecting 4 flowers: {2, 2}.

Sample 3. There are three ways of selecting 5 flowers: {1, 2, 2}, {0, 3, 2}, and {1, 3, 1}.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const long double EPS = 1e-10;
const long long int MOD = 1000000007ll;
const long long int mod1 = 1000000009ll;
const long long int mod2 = 1100000009ll;
int INF = (int)1e9;
long long int INFINF = (long long int)1e18;
int debug = 0;
long double PI = acos(-1.0);
using namespace std;
long long int bit_count(long long int _x) {
  long long int _ret = 0;
  while (_x) {
    if (_x % 2 == 1) _ret++;
    _x /= 2;
  }
  return _ret;
}
long long int bit(long long int _mask, long long int _i) {
  return (_mask & (1ll << _i)) == 0 ? 0 : 1;
}
long long int powermod(long long int _a, long long int _b,
                       long long int _m = MOD) {
  long long int _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a) % _m;
    _b /= 2;
    _a = (_a * _a) % _m;
  }
  return _r;
}
long long int add(long long int a, long long int b, long long int m = MOD) {
  long long int x = a + b;
  while (x >= m) x -= m;
  while (x < 0) x += m;
  return x;
}
long long int sub(long long int a, long long int b, long long int m = MOD) {
  long long int x = a - b;
  while (x < 0) x += m;
  while (x >= m) x -= m;
  return x;
}
long long int mul(long long int a, long long int b, long long int m = MOD) {
  long long int x = a * 1ll * b;
  x %= m;
  while (x < 0) x += m;
  return x;
}
long long int N;
long long int second;
long long int first[25];
long long int ans;
long long int inv[25];
long long int value(long long int n, long long int r) {
  long long int ret = 1;
  for (long long int a = 1, b = n; a <= r; a++, b--) {
    ret = mul(ret, b % MOD);
    ret = mul(ret, inv[a]);
  }
  return ret;
}
void solve(long long int idx, long long int coeff, long long int P) {
  if (idx == N + 1) {
    if (P > second) return;
    long long int rem = second - P;
    ans = add(ans, mul(coeff, value(N + rem - 1, N - 1)));
    return;
  }
  solve(idx + 1, coeff, P);
  solve(idx + 1, coeff * -1, P + first[idx] + 1);
}
int main() {
  if (0) {
    freopen("input.txt", "r", stdin);
    debug = 1;
  }
  srand(time(NULL));
  for (int i = 1; i < 25; i++) inv[i] = powermod(i, MOD - 2, MOD);
  scanf("%lld", &N);
  ;
  scanf("%lld", &second);
  ;
  for (long long int i = 1; i <= N; i++) scanf("%lld", &first[i]);
  ;
  solve(1, 1, 0);
  cout << ans;
  return 0;
}