#include <bits/stdc++.h>
using namespace std;
long long n, m;
struct matrix {
  long long a[5][5];
  long long n, m;
  matrix() {
    n = m = 0;
    memset(a, 0x3f, sizeof(a));
  }
  inline long long *operator[](long long x) { return a[x]; }
  matrix operator*(matrix b) {
    matrix c;
    c.n = n;
    c.m = b.m;
    for (long long i = 0; i < c.n; ++i)
      for (long long j = 0; j < c.m; ++j)
        for (long long k = 0; k < m; ++k)
          c.a[i][j] = min(c.a[i][j], a[i][k] + b.a[k][j]);
    return c;
  }
};
struct segment_tree {
  long long l, r;
  matrix k;
} tr[200010 << 2];
long long read() {
  long long s = 0, w = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') w = -1, c = getchar();
  while (c <= '9' && c >= '0') s = (s << 3) + (s << 1) + c - '0', c = getchar();
  return s * w;
}
char s[200010] = {};
inline void push(long long p) { tr[p].k = tr[p << 1].k * tr[p << 1 | 1].k; }
void build(long long p, long long l, long long r) {
  tr[p].l = l;
  tr[p].r = r;
  if (l == r) {
    tr[p].k.n = tr[p].k.m = 5;
    for (long long i = 0; i < 5; ++i) tr[p].k[i][i] = 0;
    if (s[l] == '2') tr[p].k[0][0] = 1, tr[p].k[0][1] = 0;
    if (s[l] == '0') tr[p].k[1][1] = 1, tr[p].k[1][2] = 0;
    if (s[l] == '1') tr[p].k[2][2] = 1, tr[p].k[2][3] = 0;
    if (s[l] == '7') tr[p].k[3][3] = 1, tr[p].k[3][4] = 0;
    if (s[l] == '6') tr[p].k[3][3] = 1, tr[p].k[4][4] = 1;
    return;
  }
  long long mid = (l + r) >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  push(p);
}
matrix ask(long long p, long long l, long long r) {
  if (l <= tr[p].l && tr[p].r <= r) return tr[p].k;
  long long mid = (tr[p].l + tr[p].r) >> 1;
  if (r <= mid) return ask(p << 1, l, r);
  if (l > mid) return ask(p << 1 | 1, l, r);
  return ask(p << 1, l, r) * ask(p << 1 | 1, l, r);
}
signed main() {
  n = read();
  m = read();
  cin >> (s + 1);
  build(1, 1, n);
  for (long long i = 1; i <= m; ++i) {
    long long x = read(), y = read();
    matrix ans = ask(1, x, y);
    printf("%lld\n", ans[0][4] > n ? -1 : ans[0][4]);
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
long long n, m;
struct matrix {
  long long a[5][5];
  long long n, m;
  matrix() {
    n = m = 0;
    memset(a, 0x3f, sizeof(a));
  }
  inline long long *operator[](long long x) { return a[x]; }
  matrix operator*(matrix b) {
    matrix c;
    c.n = n;
    c.m = b.m;
    for (long long i = 0; i < c.n; ++i)
      for (long long j = 0; j < c.m; ++j)
        for (long long k = 0; k < m; ++k)
          c.a[i][j] = min(c.a[i][j], a[i][k] + b.a[k][j]);
    return c;
  }
};
struct segment_tree {
  long long l, r;
  matrix k;
} tr[200010 << 2];
long long read() {
  long long s = 0, w = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') w = -1, c = getchar();
  while (c <= '9' && c >= '0') s = (s << 3) + (s << 1) + c - '0', c = getchar();
  return s * w;
}
char s[200010] = {};
inline void push(long long p) { tr[p].k = tr[p << 1].k * tr[p << 1 | 1].k; }
void build(long long p, long long l, long long r) {
  tr[p].l = l;
  tr[p].r = r;
  if (l == r) {
    tr[p].k.n = tr[p].k.m = 5;
    for (long long i = 0; i < 5; ++i) tr[p].k[i][i] = 0;
    if (s[l] == '2') tr[p].k[0][0] = 1, tr[p].k[0][1] = 0;
    if (s[l] == '0') tr[p].k[1][1] = 1, tr[p].k[1][2] = 0;
    if (s[l] == '1') tr[p].k[2][2] = 1, tr[p].k[2][3] = 0;
    if (s[l] == '7') tr[p].k[3][3] = 1, tr[p].k[3][4] = 0;
    if (s[l] == '6') tr[p].k[3][3] = 1, tr[p].k[4][4] = 1;
    return;
  }
  long long mid = (l + r) >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  push(p);
}
matrix ask(long long p, long long l, long long r) {
  if (l <= tr[p].l && tr[p].r <= r) return tr[p].k;
  long long mid = (tr[p].l + tr[p].r) >> 1;
  if (r <= mid) return ask(p << 1, l, r);
  if (l > mid) return ask(p << 1 | 1, l, r);
  return ask(p << 1, l, r) * ask(p << 1 | 1, l, r);
}
signed main() {
  n = read();
  m = read();
  cin >> (s + 1);
  build(1, 1, n);
  for (long long i = 1; i <= m; ++i) {
    long long x = read(), y = read();
    matrix ans = ask(1, x, y);
    printf("%lld\n", ans[0][4] > n ? -1 : ans[0][4]);
  }
  return 0;
}

```