#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3F3F3F3F;
const int NEGINF = 0xC0C0C0C0;
const int NULO = -1;
const double EPS = 1e-8;
using vi = vector<int>;
using vd = vector<double>;
using vvi = vector<vi>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
int cmp(double x, double y = 0, double tol = EPS) {
  return (x <= y + tol) ? (x + tol < y) ? -1 : 0 : 1;
}
template <typename T>
struct seg_tree {
  int S;
  T zero;
  vector<T> value;
  seg_tree<T>(int _S, T _zero = T()) {
    S = _S, zero = _zero;
    value.resize(2 * S + 1, zero);
  }
  void set_leaves(vector<T>& leaves) {
    copy(leaves.begin(), leaves.end(), value.begin() + S);
    for (int i = S - 1; i > 0; i--) value[i] = value[2 * i] + value[2 * i + 1];
  }
  void upd(int i, T v) {
    i += S;
    value[i] = v;
    while (i > 1) {
      i /= 2;
      value[i] = value[2 * i] + value[2 * i + 1];
    }
  }
  T query(int i, int j) {
    T res_left = zero, res_right = zero;
    for (i += S, j += S; i <= j; i /= 2, j /= 2) {
      if ((i & 1) == 1) res_left = res_left + value[i++];
      if ((j & 1) == 0) res_right = value[j--] + res_right;
    }
    return res_left + res_right;
  }
};
struct matrix {
  int mat[5][5];
  matrix(char c = '4') {
    for (__decltype(0) i = (0); i < (5); ++i)
      for (__decltype(0) j = (0); j < (5); ++j) mat[i][j] = INT_MAX / 2;
    for (__decltype(0) i = (0); i < (5); ++i) mat[i][i] = 0;
    if (c == '2') {
      mat[0][0] = 1;
      mat[0][1] = 0;
    }
    if (c == '0') {
      mat[1][1] = 1;
      mat[1][2] = 0;
    }
    if (c == '1') {
      mat[2][2] = 1;
      mat[2][3] = 0;
    }
    if (c == '7') {
      mat[3][3] = 1;
      mat[3][4] = 0;
    }
    if (c == '6') {
      mat[3][3] = 1;
      mat[4][4] = 1;
    }
  }
  matrix operator+(const matrix& rhs) const {
    matrix ans;
    for (__decltype(0) i = (0); i < (5); ++i) ans.mat[i][i] = INT_MAX / 2;
    for (__decltype(0) i = (0); i < (5); ++i)
      for (__decltype(0) j = (0); j < (5); ++j)
        for (__decltype(0) k = (0); k < (5); ++k)
          ans.mat[i][j] =
              min({ans.mat[i][j], mat[i][k] + rhs.mat[k][j], INT_MAX / 2});
    return ans;
  }
};
int main() {
  int N, Q;
  scanf("%d %d", &N, &Q);
  seg_tree<matrix> S(N);
  vector<matrix> leaves(N);
  for (int i = 0; i < N; i++) {
    char c;
    scanf(" %c", &c);
    leaves[i] = matrix(c);
  }
  S.set_leaves(leaves);
  while (Q--) {
    int a, b;
    scanf("%d %d", &a, &b);
    int ans = S.query(a - 1, b - 1).mat[0][4];
    if (ans == INT_MAX / 2) ans = -1;
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn't occur in t as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is  - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits '0'–'9'.

The i-th of next q lines contains two integers ai and bi (1 ≤ ai ≤ bi ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

Input

8 3
20166766
1 8
1 7
2 8


Output

4
3
-1


Input

15 5
012016662091670
3 4
1 14
4 15
1 13
10 15


Output

-1
2
1
-1
-1


Input

4 2
1234
2 4
1 2


Output

-1
-1

Note

In the first sample:

  * In the first query, ugliness("20166766") = 4 because all four sixes must be removed. 
  * In the second query, ugliness("2016676") = 3 because all three sixes must be removed. 
  * In the third query, ugliness("0166766") = - 1 because it's impossible to remove some digits to get a nice string. 



In the second sample:

  * In the second query, ugliness("01201666209167") = 2. It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice. 
  * In the third query, ugliness("016662091670") = 1. It's optimal to remove the last digit '6', what gives a nice string "01666209170". 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3F3F3F3F;
const int NEGINF = 0xC0C0C0C0;
const int NULO = -1;
const double EPS = 1e-8;
using vi = vector<int>;
using vd = vector<double>;
using vvi = vector<vi>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
int cmp(double x, double y = 0, double tol = EPS) {
  return (x <= y + tol) ? (x + tol < y) ? -1 : 0 : 1;
}
template <typename T>
struct seg_tree {
  int S;
  T zero;
  vector<T> value;
  seg_tree<T>(int _S, T _zero = T()) {
    S = _S, zero = _zero;
    value.resize(2 * S + 1, zero);
  }
  void set_leaves(vector<T>& leaves) {
    copy(leaves.begin(), leaves.end(), value.begin() + S);
    for (int i = S - 1; i > 0; i--) value[i] = value[2 * i] + value[2 * i + 1];
  }
  void upd(int i, T v) {
    i += S;
    value[i] = v;
    while (i > 1) {
      i /= 2;
      value[i] = value[2 * i] + value[2 * i + 1];
    }
  }
  T query(int i, int j) {
    T res_left = zero, res_right = zero;
    for (i += S, j += S; i <= j; i /= 2, j /= 2) {
      if ((i & 1) == 1) res_left = res_left + value[i++];
      if ((j & 1) == 0) res_right = value[j--] + res_right;
    }
    return res_left + res_right;
  }
};
struct matrix {
  int mat[5][5];
  matrix(char c = '4') {
    for (__decltype(0) i = (0); i < (5); ++i)
      for (__decltype(0) j = (0); j < (5); ++j) mat[i][j] = INT_MAX / 2;
    for (__decltype(0) i = (0); i < (5); ++i) mat[i][i] = 0;
    if (c == '2') {
      mat[0][0] = 1;
      mat[0][1] = 0;
    }
    if (c == '0') {
      mat[1][1] = 1;
      mat[1][2] = 0;
    }
    if (c == '1') {
      mat[2][2] = 1;
      mat[2][3] = 0;
    }
    if (c == '7') {
      mat[3][3] = 1;
      mat[3][4] = 0;
    }
    if (c == '6') {
      mat[3][3] = 1;
      mat[4][4] = 1;
    }
  }
  matrix operator+(const matrix& rhs) const {
    matrix ans;
    for (__decltype(0) i = (0); i < (5); ++i) ans.mat[i][i] = INT_MAX / 2;
    for (__decltype(0) i = (0); i < (5); ++i)
      for (__decltype(0) j = (0); j < (5); ++j)
        for (__decltype(0) k = (0); k < (5); ++k)
          ans.mat[i][j] =
              min({ans.mat[i][j], mat[i][k] + rhs.mat[k][j], INT_MAX / 2});
    return ans;
  }
};
int main() {
  int N, Q;
  scanf("%d %d", &N, &Q);
  seg_tree<matrix> S(N);
  vector<matrix> leaves(N);
  for (int i = 0; i < N; i++) {
    char c;
    scanf(" %c", &c);
    leaves[i] = matrix(c);
  }
  S.set_leaves(leaves);
  while (Q--) {
    int a, b;
    scanf("%d %d", &a, &b);
    int ans = S.query(a - 1, b - 1).mat[0][4];
    if (ans == INT_MAX / 2) ans = -1;
    printf("%d\n", ans);
  }
  return 0;
}

```