Problem: JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.

First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.

For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.

However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.

All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).

The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.

Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.

Output

Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.

Examples

Input

4 2
1011
1 4
3 4


Output

14
3


Input

3 2
111
1 2
3 3


Output

3
1

Note

In the first example: 

  * For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2. 
  * For query 2: Both 3, 4 and 4, 3 ordering give the same answer. 



In the second example, any order of eating parts leads to the same answer.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int maxn = 1e6 + 10, maxm = 2e6 + 10;
const int INF = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const double PI = acos(-1.0);
int casn, n, m, k;
int num[maxn];
int cnt[maxn];
long long p = 1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a * gcd(a, b) / b; }
long long gcd_mod(long long a, long long b, long long c = p) {
  while (b) b = ((a % b) % c + ((a = b) - b) % c) % c;
  return a;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) return (x = 1, y = 0, a);
  if (a == 0) return (x = 0, y = 1, b);
  long long r = exgcd(b, a % b, y, x);
  y -= (a / b) * x;
  return r;
}
long long lcm_mod(long long a, long long b, long long c = p) {
  return (a / gcd_mod(a, b) * b) % c;
}
long long pow_mod(long long a, long long b, long long c = p,
                  long long ans = 1) {
  while (b) {
    if (b & 1) ans = (a * ans) % c;
    a = (a * a) % c, b >>= 1;
  }
  return ans;
}
long long inv_mod(long long a, long long b = p) { return pow_mod(a, b - 2); }
long long inv_gcd(long long a, long long b = p) {
  long long x, y;
  exgcd(a, b, x, y);
  x %= b;
  while (x < 0) x += b;
  return x;
}
long long mul_mod(long long a, long long b, long long c = p) {
  if (a < b) swap(a, b);
  long long res = 0;
  while (b) {
    if (b & 1) res = (res + a) % n;
    a = (a + a) % n;
    b >>= 1;
  }
  return res;
}
long long pow_mul_mod(long long a, long long b, long long n = p) {
  long long res = 1;
  while (b) {
    if (b & 1) res = mul_mod(res, a, n);
    a = mul_mod(a, a, n);
    b >>= 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  ;
  cin >> n >> m;
  string s;
  cin >> s;
  for (int i = 1; i <= n; ++i) {
    cnt[i] = cnt[i - 1] + (s[i - 1] == '1');
  }
  while (m--) {
    int a, b;
    long long ans = 0;
    cin >> a >> b;
    if (cnt[b] - cnt[a - 1] == 0) {
      cout << 0 << '\n';
      continue;
    }
    long long xx = cnt[b] - cnt[a - 1];
    ans += pow_mod(2, xx) - 1;
    if (ans == -1) ans = mod - 1;
    if (b - a + 1 != xx) {
      ans += (ans * (pow_mod(2, b - a + 1 - xx)) - ans + mod) % mod;
    }
    ans %= mod;
    cout << ans << '\n';
  }
  return 0;
}