#include <algorithm>
#include <cstdio>
#include <vector>

const int N = 200000;
const int INF = 1000000000;

int n, m, b[N], mi[N + 1 << 1];
std::vector<int> r[N];

int get_id(int l, int r)
{
    return l + r | l != r;
}

void update(int l, int r, int k, int v)
{
    if (l <= k && k <= r) {
        int id = get_id(l, r);
        mi[id] = std::min(mi[id], v);
        if (l < r) {
            int m = l + r >> 1;
            update(l, m, k, v);
            update(m + 1, r, k, v);
        }
    }
}

int query(int l, int r, int a, int b)
{
    if (b < l || r < a) {
        return INF;
    }
    if (a <= l && r <= b) {
        return mi[get_id(l, r)];
    }
    int m = l + r >> 1;
    return std::min(query(l, m, a, b), query(m + 1, r, a, b));
}

int main()
{
#ifdef LOCAL_JUDGE
    freopen("D.in", "r", stdin);
#endif
    scanf("%d", &n);
    int zeros = 0;
    for (int i = 0; i < n; ++ i) {
        scanf("%d", b + i);
        zeros += !b[i];
    }
    scanf("%d", &m);
    for (int _ = 0; _ < m; ++ _) {
        int a, b;
        scanf("%d%d", &a, &b);
        r[a - 1].push_back(b);
    }
    for (int i = 0; i < n + 1 << 1; ++ i) {
        mi[i] = INF;
    }
    update(0, n, 0, zeros);
    for (int i = 0; i < n; ++ i) {
        for (int j : r[i]) {
            update(0, n, j, query(0, n, i, j));
        }
        update(0, n, i + 1, query(0, n, i, i) + (b[i] ? 1 : -1));
    }
    printf("%d\n", query(0, n, n, n));
}

### Prompt

Please formulate 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 <vector>

const int N = 200000;
const int INF = 1000000000;

int n, m, b[N], mi[N + 1 << 1];
std::vector<int> r[N];

int get_id(int l, int r)
{
    return l + r | l != r;
}

void update(int l, int r, int k, int v)
{
    if (l <= k && k <= r) {
        int id = get_id(l, r);
        mi[id] = std::min(mi[id], v);
        if (l < r) {
            int m = l + r >> 1;
            update(l, m, k, v);
            update(m + 1, r, k, v);
        }
    }
}

int query(int l, int r, int a, int b)
{
    if (b < l || r < a) {
        return INF;
    }
    if (a <= l && r <= b) {
        return mi[get_id(l, r)];
    }
    int m = l + r >> 1;
    return std::min(query(l, m, a, b), query(m + 1, r, a, b));
}

int main()
{
#ifdef LOCAL_JUDGE
    freopen("D.in", "r", stdin);
#endif
    scanf("%d", &n);
    int zeros = 0;
    for (int i = 0; i < n; ++ i) {
        scanf("%d", b + i);
        zeros += !b[i];
    }
    scanf("%d", &m);
    for (int _ = 0; _ < m; ++ _) {
        int a, b;
        scanf("%d%d", &a, &b);
        r[a - 1].push_back(b);
    }
    for (int i = 0; i < n + 1 << 1; ++ i) {
        mi[i] = INF;
    }
    update(0, n, 0, zeros);
    for (int i = 0; i < n; ++ i) {
        for (int j : r[i]) {
            update(0, n, j, query(0, n, i, j));
        }
        update(0, n, i + 1, query(0, n, i, i) + (b[i] ? 1 : -1));
    }
    printf("%d\n", query(0, n, n, n));
}

```