#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#define ls l, m, rt << 1 
#define rs m + 1, r, rt << 1 | 1

const int Maxv = 1000000, INF = 0x3f3f3f3f;
int imin[Maxv], dp[Maxv], cost[Maxv], n, q, cnt; 
std::vector<int> T[Maxv];

void build(int l, int r, int rt){
    if (l == r) {
        imin[rt] = INF;
        return;
    }
    int m = l + r >> 1;
    build(ls);
    build(rs);
    imin[rt] = std::min(imin[rt<<1], imin[rt << 1 | 1]);
}

int query(int L, int R, int l, int r, int rt){
    if(L <= l && r <= R) {
		return imin[rt];
	}
    int m = (l + r) >> 1, res = INF;
    if(L <= m) {
    	res = std::min(res, query(L, R, ls));	
	}
    if(R > m) {
    	res = std::min(res, query(L, R, rs));
	}
    return res;
}

void update(int pos, int val, int l, int r, int rt) {
    if (l == r) {
        imin[rt] = std::min(imin[rt], val);
        return;
    }
    int m = (l + r) >> 1;
    if (pos <= m) {
    	update(pos, val, ls);
	}
    else {
    	update(pos, val, rs);
	}
    imin[rt] = std::min(imin[rt << 1], imin[rt << 1 | 1]);
}

int main()
{
    scanf("%d",&n);
    build(1, n, 1);
    for(int i = 1, j; i <= n; i++) {
        scanf("%d", &j);
        cnt += !j;
        cost[i] = j ? 1 : -1;
    }
    scanf("%d", &q);
    for(int i = 0, j, k; i < q; i++) {
        scanf("%d%d", &j, &k);
        T[j].push_back(k);
    }
    memset(dp, INF, sizeof(dp));
    dp[0] = 0;
    for (int i = 1; i <= n; i++) {
        for (int k = 0; k < T[i].size(); k++) {
			int j = T[i][k]; 
            int mi = dp[i - 1];
            mi = std::min(mi, query(std::max(i - 1,1), j, 1, n, 1));
            if (mi < dp[j]) {
                dp[j] = mi;
                update(j, mi, 1, n, 1);
            }
        }
        dp[i] = std::min(dp[i], dp[i - 1] + cost[i]); 
    }
    printf("%d\n", dp[n] + cnt);
    return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

### Response

```cpp
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#define ls l, m, rt << 1 
#define rs m + 1, r, rt << 1 | 1

const int Maxv = 1000000, INF = 0x3f3f3f3f;
int imin[Maxv], dp[Maxv], cost[Maxv], n, q, cnt; 
std::vector<int> T[Maxv];

void build(int l, int r, int rt){
    if (l == r) {
        imin[rt] = INF;
        return;
    }
    int m = l + r >> 1;
    build(ls);
    build(rs);
    imin[rt] = std::min(imin[rt<<1], imin[rt << 1 | 1]);
}

int query(int L, int R, int l, int r, int rt){
    if(L <= l && r <= R) {
		return imin[rt];
	}
    int m = (l + r) >> 1, res = INF;
    if(L <= m) {
    	res = std::min(res, query(L, R, ls));	
	}
    if(R > m) {
    	res = std::min(res, query(L, R, rs));
	}
    return res;
}

void update(int pos, int val, int l, int r, int rt) {
    if (l == r) {
        imin[rt] = std::min(imin[rt], val);
        return;
    }
    int m = (l + r) >> 1;
    if (pos <= m) {
    	update(pos, val, ls);
	}
    else {
    	update(pos, val, rs);
	}
    imin[rt] = std::min(imin[rt << 1], imin[rt << 1 | 1]);
}

int main()
{
    scanf("%d",&n);
    build(1, n, 1);
    for(int i = 1, j; i <= n; i++) {
        scanf("%d", &j);
        cnt += !j;
        cost[i] = j ? 1 : -1;
    }
    scanf("%d", &q);
    for(int i = 0, j, k; i < q; i++) {
        scanf("%d%d", &j, &k);
        T[j].push_back(k);
    }
    memset(dp, INF, sizeof(dp));
    dp[0] = 0;
    for (int i = 1; i <= n; i++) {
        for (int k = 0; k < T[i].size(); k++) {
			int j = T[i][k]; 
            int mi = dp[i - 1];
            mi = std::min(mi, query(std::max(i - 1,1), j, 1, n, 1));
            if (mi < dp[j]) {
                dp[j] = mi;
                update(j, mi, 1, n, 1);
            }
        }
        dp[i] = std::min(dp[i], dp[i - 1] + cost[i]); 
    }
    printf("%d\n", dp[n] + cnt);
    return 0;
}

```