#include <bits/stdc++.h>
const int MAXN = 200010;
struct Tree {
  int l, r;
  int dp[6][6];
};
Tree seg[MAXN * 4];
char s[MAXN];
int trans[6][10], a[MAXN];
void ck(int &x, int y) {
  if (x == -1 || x > y) {
    x = y;
  }
}
void Up(Tree &cur, Tree &l, Tree &r) {
  memset(cur.dp, -1, sizeof(cur.dp));
  for (int i = 0; i < 6; i++) {
    for (int j = 0; j < 6; j++) {
      if (l.dp[i][j] == -1) {
        continue;
      }
      for (int k = 0; k < 6; k++) {
        if (r.dp[j][k] == -1) {
          continue;
        }
        ck(cur.dp[i][k], l.dp[i][j] + r.dp[j][k]);
      }
    }
  }
}
void build(int id, int L, int R) {
  seg[id].l = L;
  seg[id].r = R;
  if (L == R) {
    memset(seg[id].dp, -1, sizeof(seg[id].dp));
    int val = a[L], ne_state;
    for (int i = 0; i < 6; i++) {
      int ne_state;
      ne_state = trans[i][val];
      ck(seg[id].dp[i][ne_state], 0);
      ne_state = i;
      ck(seg[id].dp[i][i], 1);
    }
  } else {
    int mid = (L + R) >> 1;
    build(id * 2, L, mid);
    build(id * 2 + 1, mid + 1, R);
    Up(seg[id], seg[id * 2], seg[id * 2 + 1]);
  }
}
Tree query(int id, int L, int R) {
  if (seg[id].l >= L && seg[id].r <= R) {
    return seg[id];
  }
  int mid = (seg[id].l + seg[id].r) >> 1;
  if (R <= mid) {
    return query(id * 2, L, R);
  } else if (L > mid) {
    return query(id * 2 + 1, L, R);
  } else {
    Tree l = query(id * 2, L, mid), r = query(id * 2 + 1, mid + 1, R), res;
    Up(res, l, r);
    return res;
  }
}
int main() {
  for (int i = 0; i < 6; i++) {
    for (int j = 0; j < 10; j++) {
      trans[i][j] = i;
    }
  }
  trans[0][2] = 1;
  trans[1][0] = 2;
  trans[2][1] = 3;
  trans[3][6] = 4;
  trans[3][7] = 5;
  trans[5][6] = 6;
  int n, q;
  scanf("%d%d", &n, &q);
  scanf("%s", s);
  for (int i = 0; i < n; i++) {
    a[i] = s[i] - '0';
  }
  build(1, 0, n - 1);
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    Tree res = query(1, l, r);
    printf("%d\n", res.dp[0][5]);
  }
  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>
const int MAXN = 200010;
struct Tree {
  int l, r;
  int dp[6][6];
};
Tree seg[MAXN * 4];
char s[MAXN];
int trans[6][10], a[MAXN];
void ck(int &x, int y) {
  if (x == -1 || x > y) {
    x = y;
  }
}
void Up(Tree &cur, Tree &l, Tree &r) {
  memset(cur.dp, -1, sizeof(cur.dp));
  for (int i = 0; i < 6; i++) {
    for (int j = 0; j < 6; j++) {
      if (l.dp[i][j] == -1) {
        continue;
      }
      for (int k = 0; k < 6; k++) {
        if (r.dp[j][k] == -1) {
          continue;
        }
        ck(cur.dp[i][k], l.dp[i][j] + r.dp[j][k]);
      }
    }
  }
}
void build(int id, int L, int R) {
  seg[id].l = L;
  seg[id].r = R;
  if (L == R) {
    memset(seg[id].dp, -1, sizeof(seg[id].dp));
    int val = a[L], ne_state;
    for (int i = 0; i < 6; i++) {
      int ne_state;
      ne_state = trans[i][val];
      ck(seg[id].dp[i][ne_state], 0);
      ne_state = i;
      ck(seg[id].dp[i][i], 1);
    }
  } else {
    int mid = (L + R) >> 1;
    build(id * 2, L, mid);
    build(id * 2 + 1, mid + 1, R);
    Up(seg[id], seg[id * 2], seg[id * 2 + 1]);
  }
}
Tree query(int id, int L, int R) {
  if (seg[id].l >= L && seg[id].r <= R) {
    return seg[id];
  }
  int mid = (seg[id].l + seg[id].r) >> 1;
  if (R <= mid) {
    return query(id * 2, L, R);
  } else if (L > mid) {
    return query(id * 2 + 1, L, R);
  } else {
    Tree l = query(id * 2, L, mid), r = query(id * 2 + 1, mid + 1, R), res;
    Up(res, l, r);
    return res;
  }
}
int main() {
  for (int i = 0; i < 6; i++) {
    for (int j = 0; j < 10; j++) {
      trans[i][j] = i;
    }
  }
  trans[0][2] = 1;
  trans[1][0] = 2;
  trans[2][1] = 3;
  trans[3][6] = 4;
  trans[3][7] = 5;
  trans[5][6] = 6;
  int n, q;
  scanf("%d%d", &n, &q);
  scanf("%s", s);
  for (int i = 0; i < n; i++) {
    a[i] = s[i] - '0';
  }
  build(1, 0, n - 1);
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    Tree res = query(1, l, r);
    printf("%d\n", res.dp[0][5]);
  }
  return 0;
}

```