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;
const int INFL = (int)1e9;
const long long int INFLL = (long long int)1e18;
const double INFD = numeric_limits<double>::infinity();
const double PI = 3.14159265358979323846;
bool nearlyeq(double x, double y) { return abs(x - y) < 1e-9; }
bool inrange(int x, int t) { return x >= 0 && x < t; }
long long int rndf(double x) {
  return (long long int)(x + (x >= 0 ? 0.5 : -0.5));
}
long long int floorsqrt(double x) {
  long long int m = (long long int)sqrt(x);
  return m + (m * m <= (long long int)(x) ? 0 : -1);
}
long long int ceilsqrt(double x) {
  long long int m = (long long int)sqrt(x);
  return m + ((long long int)x <= m * m ? 0 : 1);
}
long long int rnddiv(long long int a, long long int b) {
  return (a / b + (a % b * 2 >= b ? 1 : 0));
}
long long int ceildiv(long long int a, long long int b) {
  return (a / b + (a % b == 0 ? 0 : 1));
}
long long int gcd(long long int m, long long int n) {
  if (n == 0)
    return m;
  else
    return gcd(n, m % n);
}
namespace mod_op {
const long long int MOD = (long long int)1e9 + 7;
class modll {
 private:
  long long int val;
  inline long long int modify(long long int x) {
    long long int ret = x % MOD;
    if (ret < 0) ret += MOD;
    return ret;
  }
  inline long long int inv(long long int x) {
    if (x == 0)
      return 1 / x;
    else if (x == 1)
      return 1;
    else
      return modify(inv(MOD % x) * modify(-MOD / x));
  }

 public:
  modll(long long int init = 0) {
    val = modify(init);
    return;
  }
  modll(const modll &another) {
    val = another.val;
    return;
  }
  inline modll &operator=(const modll &another) {
    val = another.val;
    return *this;
  }
  inline modll operator+(const modll &x) { return modify(val + x.val); }
  inline modll operator-(const modll &x) { return modify(val - x.val); }
  inline modll operator*(const modll &x) { return modify(val * x.val); }
  inline modll operator/(const modll &x) { return modify(val * inv(x.val)); }
  inline modll &operator+=(const modll &x) {
    val = modify(val + x.val);
    return *this;
  }
  inline modll &operator-=(const modll &x) {
    val = modify(val - x.val);
    return *this;
  }
  inline modll &operator*=(const modll &x) {
    val = modify(val * x.val);
    return *this;
  }
  inline modll &operator/=(const modll &x) {
    val = modify(val * inv(x.val));
    return *this;
  }
  inline bool operator==(const modll &x) { return val == x.val; }
  inline bool operator!=(const modll &x) { return val != x.val; }
  friend inline istream &operator>>(istream &is, modll &x) {
    is >> x.val;
    return is;
  }
  friend inline ostream &operator<<(ostream &os, const modll &x) {
    os << x.val;
    return os;
  }
  long long int get_val() { return val; }
};
modll pow(modll n, long long int p) {
  modll ret;
  if (p == 0)
    ret = 1;
  else if (p == 1)
    ret = n;
  else {
    ret = pow(n, p / 2);
    ret *= ret;
    if (p % 2 == 1) ret *= n;
  }
  return ret;
}
vector<modll> facts;
inline void make_facts(int n) {
  if (facts.empty()) facts.push_back(modll(1));
  for (int i = (int)facts.size(); i <= n; ++i)
    facts.push_back(modll(facts.back() * (long long int)i));
  return;
}
vector<modll> ifacts;
vector<modll> invs;
inline void make_invs(int n) {
  if (invs.empty()) {
    invs.push_back(modll(0));
    invs.push_back(modll(1));
  }
  for (int i = (int)invs.size(); i <= n; ++i) {
    invs.push_back(invs[(int)MOD % i] * ((int)MOD - (int)MOD / i));
  }
  return;
}
inline void make_ifacts(int n) {
  make_invs(n);
  if (ifacts.empty()) ifacts.push_back(modll(1));
  for (int i = (int)ifacts.size(); i <= n; ++i)
    ifacts.push_back(modll(ifacts.back() * invs[i]));
  return;
}
modll combination(long long int n, long long int r) {
  if (n >= r && r >= 0) {
    modll ret;
    make_facts((int)n);
    make_ifacts((int)n);
    ret = facts[(unsigned)n] * ifacts[(unsigned)r] * ifacts[(unsigned)(n - r)];
    return ret;
  } else
    return 0;
}
modll get_fact(long long int n) {
  make_facts((int)n);
  return facts[(int)n];
}
modll get_ifact(long long int n) {
  make_ifacts((int)n);
  return ifacts[(int)n];
}
long long int disc_log(modll a, modll b) {
  long long int ret = -1;
  long long int m = ceilsqrt(MOD);
  unordered_map<long long int, long long int> mp;
  modll x = 1;
  for (int i = 0; i < (int)m; i++) {
    mp[x.get_val()] = i;
    x *= a;
  }
  x = modll(1) / pow(a, m);
  modll k = b;
  for (int i = 0; i < (int)m; i++) {
    if (mp.find(k.get_val()) == mp.end())
      k *= x;
    else {
      ret = i * m + mp[k.get_val()];
      break;
    }
  }
  return ret;
}
}  // namespace mod_op
using namespace mod_op;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  int n;
  cin >> n;
  int q;
  cin >> q;
  string s;
  cin >> s;
  vector<long long int> sum(n + 1, 0);
  for (int i = 0; i < (int)n; i++) sum[i + 1] = sum[i] + (s[i] == '1' ? 1 : 0);
  for (int unused = 0; unused < (int)q; unused++) {
    int s, t;
    cin >> s >> t;
    long long int m = t - s + 1;
    long long int cnt1 = sum[t] - sum[s - 1];
    long long int cnt0 = m - cnt1;
    modll buf = pow(modll(2), cnt1) - 1;
    modll buf2 = pow(modll(2), cnt0) - 1;
    modll ans = buf + buf * buf2;
    cout << ans << "\n";
  }
}