#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
struct mint {
  long long x;
  mint(long long x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const { return mint(*this) += a; }
  mint operator-(const mint a) const { return mint(*this) -= a; }
  mint operator*(const mint a) const { return mint(*this) *= a; }
  mint pow(long long t) const {
    if (!t) return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }
  mint inv() const { return pow(mod - 2); }
  mint& operator/=(const mint a) { return *this *= a.inv(); }
  mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
mint f(int n) {
  if (n == 0) return 1;
  mint x = f(n / 2);
  x *= x;
  if (n % 2 == 1) x *= 2;
  return x;
}
long long modpow(long long a, long long n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
template <typename T>
struct BIT {
  int n;
  vector<T> d;
  BIT(int n = 0) { init(n); }
  void init(int nn) {
    n = nn;
    d.resize(nn + 1);
  }
  void add(int i, T x = 1) {
    for (i++; i <= n; i += i & -i) {
      d[i] += x;
    }
  }
  T sum(int i) {
    T x = 0;
    for (i++; i; i -= i & -i) {
      x += d[i];
    }
    return x;
  }
  T sum(int l, int r) { return sum(r - 1) - sum(l - 1); }
};
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < (n); i++) cin >> a[i];
  BIT<mint> dp[n + 1];
  for (int i = 0; i < (n + 1); i++) dp[i].init(k + 2);
  dp[0].add(k + 1, 1);
  for (int i = 0; i < (n + 1); i++) {
    for (int j = 1; j <= k + 1; j++) {
      dp[i + 1].add(j, dp[i].sum(j, min(k + 1, j + a[i]) + 1));
    }
  }
  cout << dp[n].sum(1) << endl;
}
