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>
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '{';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << '{';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T, size_t size>
ostream &operator<<(ostream &os, const array<T, size> &arr) {
  os << '{';
  string sep;
  for (const auto &x : arr) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const map<A, B> &v) {
  os << '{';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
template <class T>
bool cmin(T &a, const T &b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool cmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const long long mod = 1e9 + 7;
const long long N = 25;
long long n, s, f[20], ifact[N + 1];
long long power(long long a, long long b, long long mod) {
  long long ret = 1;
  a %= mod;
  while (b > 0) {
    if (b % 2) ret = ret * a % mod;
    b /= 2;
    a = a * a % mod;
  }
  return ret;
}
long long inv_mod(long long x) { return power(x, mod - 2, mod); }
long long get_NCR(long long N, long long R) {
  if (N < 0 || R > N) return 0;
  long long ans = 1;
  for (long long i = N; i >= (N - R + 1); i--) {
    long long mul = i % mod;
    ans = (ans * mul) % mod;
  }
  ans = (ans * ifact[R]) % mod;
  return ans;
}
void pre() {
  long long fact = 1;
  ifact[0] = inv_mod(fact);
  for (long long i = 1; i <= N; i++) {
    fact *= i;
    fact %= mod;
    ifact[i] = inv_mod(fact);
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  pre();
  cin >> n >> s;
  for (long long i = 0; i < n; i++) {
    cin >> f[i];
  }
  long long ans = 0;
  for (long long mask = 0; mask < (1 << n); mask++) {
    long long sign = 1;
    long long taken = 0;
    for (long long i = 0; i < n; i++)
      if (mask >> i & 1) {
        sign *= -1;
        taken += f[i] + 1;
      }
    long long remain = s - taken;
    long long coeff = get_NCR(n + remain - 1, n - 1);
    if (sign > 0)
      ans = (ans + coeff) % mod;
    else
      ans = (ans - coeff + mod) % mod;
  }
  cout << ans << "\n";
}