#include <bits/stdc++.h>
using namespace std;
int const lim = 2e5 + 3;
int const MLOG = 21;
int inp[lim + 3];
char zz[lim + 3];
struct nod {
  int arr[5][5];
  nod() {
    for (int i = 0; i <= 4; i++)
      for (int j = 0; j <= 4; j++) arr[i][j] = INT_MAX / 2;
  }
} st[4 * lim + 3];
int test[] = {2, 0, 1, 6};
nod merge(nod a, nod b) {
  nod ret;
  ret = nod();
  for (int i = 0; i < 4; i++)
    for (int j = i; j < 4; j++)
      for (int k = i; k <= j; k++)
        ret.arr[i][j] = min(ret.arr[i][j], a.arr[i][k] + b.arr[k][j]);
  return ret;
}
void build(int i, int j, int index) {
  if (i == j) {
    for (int k = 0; k < 4; k++) st[index].arr[k][k] = 0;
    for (int k = 0; k < 4; k++)
      if (inp[i] == test[k])
        st[index].arr[k][k + 1] = 0, st[index].arr[k][k] = 1;
    return;
  }
  int mid = (i + j) / 2;
  build(i, mid, index * 2 + 1);
  build(mid + 1, j, index * 2 + 2);
  st[index] = merge(st[index * 2 + 1], st[index * 2 + 2]);
}
nod get(int i, int j, int index, int l, int r) {
  if (i > r || j < l) {
    nod ret;
    for (int i = 0; i < 4; i++)
      for (int j = 0; j < 4; j++) ret.arr[i][j] = 0;
    return ret;
  }
  if (i >= l && j <= r) return st[index];
  int mid = (i + j) / 2;
  nod val1 = get(i, mid, index * 2 + 1, l, r);
  nod val2 = get(mid + 1, j, index * 2 + 2, l, r);
  if (mid + 1 > r) return val1;
  if (mid < l) return val2;
  return merge(val1, val2);
}
int sum[lim + 3], last[lim + 3];
int main() {
  int n, q;
  cin >> n >> q;
  char x;
  scanf("%c", &x);
  for (int i = 1; i <= n; i++) {
    scanf("%c", &x);
    inp[i] = x - '0';
  }
  build(1, n, 0);
  vector<int> sev, six;
  last[0] = -100;
  for (int i = 1; i <= n; i++) {
    if (inp[i] == 6)
      sum[i] = sum[i - 1] + 1;
    else
      sum[i] = sum[i - 1];
    if (inp[i] == 7)
      last[i] = i;
    else
      last[i] = last[i - 1];
  }
  while (q--) {
    int u, v;
    scanf("%d%d", &u, &v);
    if (last[v] < u) {
      puts("-1");
      continue;
    }
    int ind = last[v];
    nod res = get(1, n, 0, u, ind);
    int ans = res.arr[0][3];
    if (ans > n) {
      puts("-1");
      continue;
    }
    ans += sum[v] - sum[ind];
    printf("%d\n", ans);
  }
}

### Prompt

Create a solution in Cpp for 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;
int const lim = 2e5 + 3;
int const MLOG = 21;
int inp[lim + 3];
char zz[lim + 3];
struct nod {
  int arr[5][5];
  nod() {
    for (int i = 0; i <= 4; i++)
      for (int j = 0; j <= 4; j++) arr[i][j] = INT_MAX / 2;
  }
} st[4 * lim + 3];
int test[] = {2, 0, 1, 6};
nod merge(nod a, nod b) {
  nod ret;
  ret = nod();
  for (int i = 0; i < 4; i++)
    for (int j = i; j < 4; j++)
      for (int k = i; k <= j; k++)
        ret.arr[i][j] = min(ret.arr[i][j], a.arr[i][k] + b.arr[k][j]);
  return ret;
}
void build(int i, int j, int index) {
  if (i == j) {
    for (int k = 0; k < 4; k++) st[index].arr[k][k] = 0;
    for (int k = 0; k < 4; k++)
      if (inp[i] == test[k])
        st[index].arr[k][k + 1] = 0, st[index].arr[k][k] = 1;
    return;
  }
  int mid = (i + j) / 2;
  build(i, mid, index * 2 + 1);
  build(mid + 1, j, index * 2 + 2);
  st[index] = merge(st[index * 2 + 1], st[index * 2 + 2]);
}
nod get(int i, int j, int index, int l, int r) {
  if (i > r || j < l) {
    nod ret;
    for (int i = 0; i < 4; i++)
      for (int j = 0; j < 4; j++) ret.arr[i][j] = 0;
    return ret;
  }
  if (i >= l && j <= r) return st[index];
  int mid = (i + j) / 2;
  nod val1 = get(i, mid, index * 2 + 1, l, r);
  nod val2 = get(mid + 1, j, index * 2 + 2, l, r);
  if (mid + 1 > r) return val1;
  if (mid < l) return val2;
  return merge(val1, val2);
}
int sum[lim + 3], last[lim + 3];
int main() {
  int n, q;
  cin >> n >> q;
  char x;
  scanf("%c", &x);
  for (int i = 1; i <= n; i++) {
    scanf("%c", &x);
    inp[i] = x - '0';
  }
  build(1, n, 0);
  vector<int> sev, six;
  last[0] = -100;
  for (int i = 1; i <= n; i++) {
    if (inp[i] == 6)
      sum[i] = sum[i - 1] + 1;
    else
      sum[i] = sum[i - 1];
    if (inp[i] == 7)
      last[i] = i;
    else
      last[i] = last[i - 1];
  }
  while (q--) {
    int u, v;
    scanf("%d%d", &u, &v);
    if (last[v] < u) {
      puts("-1");
      continue;
    }
    int ind = last[v];
    nod res = get(1, n, 0, u, ind);
    int ans = res.arr[0][3];
    if (ans > n) {
      puts("-1");
      continue;
    }
    ans += sum[v] - sum[ind];
    printf("%d\n", ans);
  }
}

```