#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const int maxn = 200004;
string yr = "2017y";
char str[maxn];
struct node {
  int a[5][5];
  node() {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) a[i][j] = inf;
  }
} emp;
node seg[4 * maxn];
node Merge(node p, node q) {
  node c;
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++)
        c.a[i][j] = min(c.a[i][j], p.a[i][k] + q.a[k][j]);
    }
  }
  return c;
}
void build(int v, int l, int r) {
  if (l > r) return;
  if (l == r) {
    for (int i = 0; i < 5; i++) {
      if (str[l] == yr[i]) {
        seg[v].a[i][i + 1] = 0;
        seg[v].a[i][i] = 1;
      } else
        seg[v].a[i][i] = 0;
    }
    if (str[l] == '6') {
      seg[v].a[3][3] = 1;
      seg[v].a[4][4] = 1;
    }
    return;
  }
  int mid = (l + r) / 2;
  build(2 * v, l, mid);
  build(2 * v + 1, mid + 1, r);
  seg[v] = Merge(seg[2 * v], seg[2 * v + 1]);
}
node query(int v, int l, int r, int ll, int rr) {
  if (l > r || ll > r || rr < l) return emp;
  int mid = (l + r) / 2;
  if (ll <= l && r <= rr)
    return seg[v];
  else
    return (Merge(query(2 * v, l, mid, ll, rr),
                  query(2 * v + 1, mid + 1, r, ll, rr)));
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  scanf("%s", str + 1);
  for (int i = 0; i < 5; i++) emp.a[i][i] = 0;
  build(1, 1, n);
  int l, r;
  while (q--) {
    scanf("%d %d", &l, &r);
    node ans = query(1, 1, n, l, r);
    if (ans.a[0][4] == inf)
      printf("-1\n");
    else
      printf("%d\n", ans.a[0][4]);
  }
  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 inf = 1e9;
const int maxn = 200004;
string yr = "2017y";
char str[maxn];
struct node {
  int a[5][5];
  node() {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) a[i][j] = inf;
  }
} emp;
node seg[4 * maxn];
node Merge(node p, node q) {
  node c;
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++)
        c.a[i][j] = min(c.a[i][j], p.a[i][k] + q.a[k][j]);
    }
  }
  return c;
}
void build(int v, int l, int r) {
  if (l > r) return;
  if (l == r) {
    for (int i = 0; i < 5; i++) {
      if (str[l] == yr[i]) {
        seg[v].a[i][i + 1] = 0;
        seg[v].a[i][i] = 1;
      } else
        seg[v].a[i][i] = 0;
    }
    if (str[l] == '6') {
      seg[v].a[3][3] = 1;
      seg[v].a[4][4] = 1;
    }
    return;
  }
  int mid = (l + r) / 2;
  build(2 * v, l, mid);
  build(2 * v + 1, mid + 1, r);
  seg[v] = Merge(seg[2 * v], seg[2 * v + 1]);
}
node query(int v, int l, int r, int ll, int rr) {
  if (l > r || ll > r || rr < l) return emp;
  int mid = (l + r) / 2;
  if (ll <= l && r <= rr)
    return seg[v];
  else
    return (Merge(query(2 * v, l, mid, ll, rr),
                  query(2 * v + 1, mid + 1, r, ll, rr)));
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  scanf("%s", str + 1);
  for (int i = 0; i < 5; i++) emp.a[i][i] = 0;
  build(1, 1, n);
  int l, r;
  while (q--) {
    scanf("%d %d", &l, &r);
    node ans = query(1, 1, n, l, r);
    if (ans.a[0][4] == inf)
      printf("-1\n");
    else
      printf("%d\n", ans.a[0][4]);
  }
  return 0;
}

```