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>
using namespace std;
long long mod = 1000000007;
long long mod2 = 998244353;
long long mod3 = 1000003;
long long mod4 = 998244853;
long long mod5 = 1000000009;
long long inf = 1LL << 62;
double pi = 3.141592653589793238462643383279L;
double eps = 1e-14;
int dh[4] = {1, -1, 0, 0};
int dw[4] = {0, 0, 1, -1};
int ddh[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int ddw[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  long long c = gcd(a, b);
  return a * b / c;
}
long long Pow(long long n, long long k) {
  long long ret = 1;
  long long now = n;
  while (k > 0) {
    if (k & 1) ret *= now;
    now *= now;
    k /= 2;
  }
  return ret;
}
long long beki(long long n, long long k, long long md) {
  long long ret = 1;
  long long now = n;
  now %= md;
  while (k > 0) {
    if (k % 2 == 1) {
      ret *= now;
      ret %= md;
    }
    now *= now;
    now %= md;
    k /= 2;
  }
  return ret;
}
long long gyaku(long long n, long long md) { return beki(n, md - 2, md); }
long long popcount(long long n) {
  long long ret = 0;
  long long u = n;
  while (u > 0) {
    ret += u % 2;
    u /= 2;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  string s;
  cin >> s;
  int a[n];
  for (long long i = 0; i < n; i++) a[i] = s[i] - '0';
  int sum[n + 1];
  sum[0] = 0;
  for (long long i = 0; i < n; i++) {
    sum[i + 1] = sum[i] + a[i];
  }
  for (long long i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    long long cnt = sum[r + 1] - sum[l];
    long long ans = beki(2, r - l + 1, mod) - 1 + mod;
    cnt = r - l + 1 - cnt;
    ans += -beki(2, cnt, mod) + 1 + mod;
    ans %= mod;
    cout << ans << endl;
  }
}