//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker,"/STACK:102400000,102400000")
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha\n")
/*
定义dp(x,i,j)表示区间[0,x)内能够获得的2017的前缀长度为i的，至少要删除多少个字符
最少要删除多少个才能构成前缀为j的2017,然后我们不需要去管x，只需要用线段树去维护i，j即可。
*/
const int inf = 0x3f3f3f3f;
const int maxn = 200000 + 5;
struct Node{
    int dp[5][5];
}tree[maxn << 2];
int n, q;
char ch[maxn];
map<char, int> id;

inline void init(Node &t){
    for (int i = 0; i < 5; i++)
        for (int j = 0; j < 5; j++)
            t.dp[i][j] = inf;
}

Node Merge(Node a, Node b){
    Node tmp; init(tmp);
    for (int i = 0; i <= 4; i++)
        for (int j = i; j <= 4; j++)
            for (int k = i; k <= j; k++)
                tmp.dp[i][j] = min(tmp.dp[i][j], a.dp[i][k] + b.dp[k][j]);
    return tmp;
}

void display(Node t){
    for (int i = 0; i < 5; i++){
        for (int j = 0; j < 5; j++)
            printf("%d ", t.dp[i][j]);
        cout << endl;
    }
}

void buildtree(int l, int r, int o){
    if (l == r){
        init(tree[o]);
        for (int i = 0; i < 5; i++)
            tree[o].dp[i][i] = 0;
        int pos = id[ch[l]];
        if (pos <= 3) tree[o].dp[pos][pos] = 1, tree[o].dp[pos][pos + 1] = 0;
        if (pos == 4) tree[o].dp[3][3] = tree[o].dp[4][4] = 1;
        return ;
    }
    int mid = (l + r) / 2;
    if (l <= mid) buildtree(l, mid, o << 1);
    if (r > mid) buildtree(mid + 1, r, o << 1 | 1);
    tree[o] = Merge(tree[o << 1], tree[o << 1 | 1]);
    //display(tree[o]);
}

Node ans;
void query(int ql, int qr, int l, int r, int o){
    if (ql <= l && qr >= r) {
        if (ql == l) ans = tree[o];
        else ans = Merge(ans, tree[o]);
        return ;
    }
    int mid = (l + r) / 2;
    if (ql <= mid) query(ql, qr, l, mid, o << 1);
    if (qr > mid) query(ql, qr, mid + 1, r, o << 1 | 1);
}

int main(){
    int cnt = 4;
    id[‘2‘] = 0, id[‘0‘] = 1, id[‘1‘] = 2, id[‘7‘] = 3, id[‘6‘] = 4;
    for (char i = ‘3‘; i <= ‘9‘; i++) if(id.count(i) == 0) id[i] = ++cnt;
    cin >> n >> q;
    scanf("%s", ch + 1);
    buildtree(1, n, 1);
    for (int i = 1; i <= q; i++){
        int ql, qr; scanf("%d%d", &ql, &qr);
        init(ans);
        query(ql, qr, 1, n, 1);
        if (ans.dp[0][4] >= inf) puts("-1");
        else printf("%d\n", ans.dp[0][4]);
    }
    return 0;
}