#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200009;
struct node {
  int a[5][5];
  node() {
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) a[i][j] = (i != j) * (1 << 29);
  }
  friend node operator*(const node &x, const node &y) {
    node z;
    for (int i = 0; i < 5; ++i) z.a[i][i] = (1 << 29);
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j)
        for (int k = 0; k < 5; ++k)
          z.a[i][j] = min(z.a[i][j], x.a[i][k] + y.a[k][j]);
    return z;
  }
} tr[MAXN * 4];
int n, m;
char s[MAXN];
void build(int x, int l, int r) {
  if (l == r) {
    if (s[l] == '2') {
      tr[x].a[0][0] = 1, tr[x].a[0][1] = 0;
      return;
    }
    if (s[l] == '0') {
      tr[x].a[1][1] = 1, tr[x].a[1][2] = 0;
      return;
    }
    if (s[l] == '1') {
      tr[x].a[2][2] = 1, tr[x].a[2][3] = 0;
      return;
    }
    if (s[l] == '6') {
      tr[x].a[3][3] = tr[x].a[4][4] = 1;
      return;
    }
    if (s[l] == '7') {
      tr[x].a[3][3] = 1, tr[x].a[3][4] = 0;
      return;
    }
    return;
  }
  int mid = l + r >> 1;
  build((x << 1), l, mid), build((x << 1 | 1), mid + 1, r);
  tr[x] = tr[(x << 1)] * tr[(x << 1 | 1)];
}
node query(int x, int l, int r, int al, int ar) {
  if (l >= al && r <= ar) return tr[x];
  int mid = l + r >> 1;
  if (al <= mid && ar > mid)
    return query((x << 1), l, mid, al, ar) *
           query((x << 1 | 1), mid + 1, r, al, ar);
  if (al <= mid) return query((x << 1), l, mid, al, ar);
  return query((x << 1 | 1), mid + 1, r, al, ar);
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  build(1, 1, n);
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int ans = query(1, 1, n, l, r).a[0][4];
    printf("%d\n", ans <= n ? ans : -1);
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 = 200009;
struct node {
  int a[5][5];
  node() {
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) a[i][j] = (i != j) * (1 << 29);
  }
  friend node operator*(const node &x, const node &y) {
    node z;
    for (int i = 0; i < 5; ++i) z.a[i][i] = (1 << 29);
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j)
        for (int k = 0; k < 5; ++k)
          z.a[i][j] = min(z.a[i][j], x.a[i][k] + y.a[k][j]);
    return z;
  }
} tr[MAXN * 4];
int n, m;
char s[MAXN];
void build(int x, int l, int r) {
  if (l == r) {
    if (s[l] == '2') {
      tr[x].a[0][0] = 1, tr[x].a[0][1] = 0;
      return;
    }
    if (s[l] == '0') {
      tr[x].a[1][1] = 1, tr[x].a[1][2] = 0;
      return;
    }
    if (s[l] == '1') {
      tr[x].a[2][2] = 1, tr[x].a[2][3] = 0;
      return;
    }
    if (s[l] == '6') {
      tr[x].a[3][3] = tr[x].a[4][4] = 1;
      return;
    }
    if (s[l] == '7') {
      tr[x].a[3][3] = 1, tr[x].a[3][4] = 0;
      return;
    }
    return;
  }
  int mid = l + r >> 1;
  build((x << 1), l, mid), build((x << 1 | 1), mid + 1, r);
  tr[x] = tr[(x << 1)] * tr[(x << 1 | 1)];
}
node query(int x, int l, int r, int al, int ar) {
  if (l >= al && r <= ar) return tr[x];
  int mid = l + r >> 1;
  if (al <= mid && ar > mid)
    return query((x << 1), l, mid, al, ar) *
           query((x << 1 | 1), mid + 1, r, al, ar);
  if (al <= mid) return query((x << 1), l, mid, al, ar);
  return query((x << 1 | 1), mid + 1, r, al, ar);
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  build(1, 1, n);
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int ans = query(1, 1, n, l, r).a[0][4];
    printf("%d\n", ans <= n ? ans : -1);
  }
  return 0;
}

```