#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
const int INF = 0x3f3f3f3f;
int n, q;
char s[N];
int base[10][5][5];
inline void chmin(int &x, int y) {
  if (x > y) x = y;
}
struct SEG {
  struct Matrix {
    int a[5][5];
    Matrix() { memset(a, 0x3f, sizeof a); }
    Matrix(int _a[5][5]) {
      for (int i = 0; i < 5; ++i)
        for (int j = 0; j < 5; ++j) a[i][j] = _a[i][j];
    }
    bool isbase() {
      for (int i = 0; i < 5; ++i)
        for (int j = 0; j < 5; ++j) {
          if (i == j)
            if (a[i][j] != 1) return 0;
          if (i != j)
            if (a[i][j] != 0) return 0;
        }
      return 1;
    }
    void setbase() {
      memset(a, 0, sizeof a);
      for (int i = 0; i < 5; ++i) a[i][i] = 1;
    }
    Matrix operator*(const Matrix &other) const {
      Matrix res = Matrix();
      for (int i = 0; i < 5; ++i) {
        for (int j = 0; j < 5; ++j) {
          for (int k = 0; k < 5; ++k) {
            chmin(res.a[i][j], a[i][k] + other.a[k][j]);
          }
        }
      }
      return res;
    }
  } t[N << 2], res;
  void build(int id, int l, int r) {
    if (l == r) {
      t[id] = Matrix(base[s[l] - '0']);
      return;
    }
    int mid = (l + r) >> 1;
    build(id << 1, l, mid);
    build(id << 1 | 1, mid + 1, r);
    t[id] = t[id << 1] * t[id << 1 | 1];
  }
  void query(int id, int l, int r, int ql, int qr) {
    if (l >= ql && r <= qr) {
      res = res * t[id];
      return;
    }
    int mid = (l + r) >> 1;
    if (ql <= mid) query(id << 1, l, mid, ql, qr);
    if (qr > mid) query(id << 1 | 1, mid + 1, r, ql, qr);
  }
} seg;
int main() {
  memset(base, 0x3f, sizeof base);
  for (int i = 0; i < 10; ++i)
    for (int j = 0; j < 5; ++j) base[i][j][j] = 0;
  base[2][0][0] = 1;
  base[2][0][1] = 0;
  base[0][1][1] = 1;
  base[0][1][2] = 0;
  base[1][2][3] = 0;
  base[1][2][2] = 1;
  base[7][3][4] = 0;
  base[7][3][3] = 1;
  base[6][3][3] = 1;
  base[6][4][4] = 1;
  while (scanf("%d%d", &n, &q) != EOF) {
    scanf("%s", s + 1);
    seg.build(1, 1, n);
    for (int i = 1, l, r; i <= q; ++i) {
      scanf("%d%d", &l, &r);
      seg.res = SEG::Matrix();
      for (int i = 0; i < 5; ++i) seg.res.a[i][i] = 0;
      seg.query(1, 1, n, l, r);
      if (seg.res.a[0][4] >= INF)
        puts("-1");
      else
        printf("%d\n", seg.res.a[0][4]);
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 N = 2e5 + 10;
const int INF = 0x3f3f3f3f;
int n, q;
char s[N];
int base[10][5][5];
inline void chmin(int &x, int y) {
  if (x > y) x = y;
}
struct SEG {
  struct Matrix {
    int a[5][5];
    Matrix() { memset(a, 0x3f, sizeof a); }
    Matrix(int _a[5][5]) {
      for (int i = 0; i < 5; ++i)
        for (int j = 0; j < 5; ++j) a[i][j] = _a[i][j];
    }
    bool isbase() {
      for (int i = 0; i < 5; ++i)
        for (int j = 0; j < 5; ++j) {
          if (i == j)
            if (a[i][j] != 1) return 0;
          if (i != j)
            if (a[i][j] != 0) return 0;
        }
      return 1;
    }
    void setbase() {
      memset(a, 0, sizeof a);
      for (int i = 0; i < 5; ++i) a[i][i] = 1;
    }
    Matrix operator*(const Matrix &other) const {
      Matrix res = Matrix();
      for (int i = 0; i < 5; ++i) {
        for (int j = 0; j < 5; ++j) {
          for (int k = 0; k < 5; ++k) {
            chmin(res.a[i][j], a[i][k] + other.a[k][j]);
          }
        }
      }
      return res;
    }
  } t[N << 2], res;
  void build(int id, int l, int r) {
    if (l == r) {
      t[id] = Matrix(base[s[l] - '0']);
      return;
    }
    int mid = (l + r) >> 1;
    build(id << 1, l, mid);
    build(id << 1 | 1, mid + 1, r);
    t[id] = t[id << 1] * t[id << 1 | 1];
  }
  void query(int id, int l, int r, int ql, int qr) {
    if (l >= ql && r <= qr) {
      res = res * t[id];
      return;
    }
    int mid = (l + r) >> 1;
    if (ql <= mid) query(id << 1, l, mid, ql, qr);
    if (qr > mid) query(id << 1 | 1, mid + 1, r, ql, qr);
  }
} seg;
int main() {
  memset(base, 0x3f, sizeof base);
  for (int i = 0; i < 10; ++i)
    for (int j = 0; j < 5; ++j) base[i][j][j] = 0;
  base[2][0][0] = 1;
  base[2][0][1] = 0;
  base[0][1][1] = 1;
  base[0][1][2] = 0;
  base[1][2][3] = 0;
  base[1][2][2] = 1;
  base[7][3][4] = 0;
  base[7][3][3] = 1;
  base[6][3][3] = 1;
  base[6][4][4] = 1;
  while (scanf("%d%d", &n, &q) != EOF) {
    scanf("%s", s + 1);
    seg.build(1, 1, n);
    for (int i = 1, l, r; i <= q; ++i) {
      scanf("%d%d", &l, &r);
      seg.res = SEG::Matrix();
      for (int i = 0; i < 5; ++i) seg.res.a[i][i] = 0;
      seg.query(1, 1, n, l, r);
      if (seg.res.a[0][4] >= INF)
        puts("-1");
      else
        printf("%d\n", seg.res.a[0][4]);
    }
  }
  return 0;
}

```