#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
const int inf = 0x3f3f3f3f;
char s[maxn];
int n, m;
struct matrix {
  int mat[5][5];
  matrix() { memset(mat, 0x3f, sizeof(mat)); }
  int *operator[](int x) { return mat[x]; }
  friend matrix operator*(matrix &x, matrix &y) {
    matrix z;
    for (int i = 0; i <= 4; i++) {
      for (int j = 0; j <= 4; j++) {
        for (int k = 0; k <= 4; k++) {
          z[i][j] = min(z[i][j], x[i][k] + y[k][j]);
        }
      }
    }
    return z;
  }
};
struct node {
  int lc, rc;
  matrix mat;
} t[maxn << 1];
int tot, rt;
inline void pull(int o) { t[o].mat = t[t[o].lc].mat * t[t[o].rc].mat; }
void build(int &o, int l, int r) {
  o = ++tot;
  if (l == r) {
    for (int i = 0; i < 5; i++) t[o].mat[i][i] = 0;
    if (s[l] == '2') t[o].mat[0][1] = 0, t[o].mat[0][0] = 1;
    if (s[l] == '0') t[o].mat[1][2] = 0, t[o].mat[1][1] = 1;
    if (s[l] == '1') t[o].mat[2][3] = 0, t[o].mat[2][2] = 1;
    if (s[l] == '7') t[o].mat[3][4] = 0, t[o].mat[3][3] = 1;
    if (s[l] == '6') t[o].mat[3][3] = 1, t[o].mat[4][4] = 1;
    return;
  }
  int mid = l + r >> 1;
  build(t[o].lc, l, mid);
  build(t[o].rc, mid + 1, r);
  pull(o);
}
matrix query(int o, int l, int r, int x, int y) {
  if (l == x && r == y) {
    return t[o].mat;
  }
  int mid = l + r >> 1;
  if (y <= mid) {
    return query(t[o].lc, l, mid, x, y);
  } else if (x > mid) {
    return query(t[o].rc, mid + 1, r, x, y);
  } else {
    matrix ansl = query(t[o].lc, l, mid, x, mid);
    matrix ansr = query(t[o].rc, mid + 1, r, mid + 1, y);
    return ansl * ansr;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> m >> s + 1;
  build(rt, 1, n);
  while (m--) {
    int l, r;
    cin >> l >> r;
    int ans = query(rt, 1, n, l, r)[0][4];
    cout << (ans == inf ? -1 : ans) << endl;
  }
  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 maxn = 2e5 + 10;
const int inf = 0x3f3f3f3f;
char s[maxn];
int n, m;
struct matrix {
  int mat[5][5];
  matrix() { memset(mat, 0x3f, sizeof(mat)); }
  int *operator[](int x) { return mat[x]; }
  friend matrix operator*(matrix &x, matrix &y) {
    matrix z;
    for (int i = 0; i <= 4; i++) {
      for (int j = 0; j <= 4; j++) {
        for (int k = 0; k <= 4; k++) {
          z[i][j] = min(z[i][j], x[i][k] + y[k][j]);
        }
      }
    }
    return z;
  }
};
struct node {
  int lc, rc;
  matrix mat;
} t[maxn << 1];
int tot, rt;
inline void pull(int o) { t[o].mat = t[t[o].lc].mat * t[t[o].rc].mat; }
void build(int &o, int l, int r) {
  o = ++tot;
  if (l == r) {
    for (int i = 0; i < 5; i++) t[o].mat[i][i] = 0;
    if (s[l] == '2') t[o].mat[0][1] = 0, t[o].mat[0][0] = 1;
    if (s[l] == '0') t[o].mat[1][2] = 0, t[o].mat[1][1] = 1;
    if (s[l] == '1') t[o].mat[2][3] = 0, t[o].mat[2][2] = 1;
    if (s[l] == '7') t[o].mat[3][4] = 0, t[o].mat[3][3] = 1;
    if (s[l] == '6') t[o].mat[3][3] = 1, t[o].mat[4][4] = 1;
    return;
  }
  int mid = l + r >> 1;
  build(t[o].lc, l, mid);
  build(t[o].rc, mid + 1, r);
  pull(o);
}
matrix query(int o, int l, int r, int x, int y) {
  if (l == x && r == y) {
    return t[o].mat;
  }
  int mid = l + r >> 1;
  if (y <= mid) {
    return query(t[o].lc, l, mid, x, y);
  } else if (x > mid) {
    return query(t[o].rc, mid + 1, r, x, y);
  } else {
    matrix ansl = query(t[o].lc, l, mid, x, mid);
    matrix ansr = query(t[o].rc, mid + 1, r, mid + 1, y);
    return ansl * ansr;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> m >> s + 1;
  build(rt, 1, n);
  while (m--) {
    int l, r;
    cin >> l >> r;
    int ans = query(rt, 1, n, l, r)[0][4];
    cout << (ans == inf ? -1 : ans) << endl;
  }
  return 0;
}

```