#include <bits/stdc++.h>

using namespace std;

const int INF = 2e9;

struct Node {
    typedef Node* NP;
    NP l, r;
    int sz, mi;
    Node(int _sz) : l(NULL), r(NULL), sz(_sz) {
        mi = INF;
        if (sz == 1) return;
        l = new Node(sz/2);
        r = new Node(sz-sz/2);
    }
    void set(int k, int v) {
        if (sz == 1) {
            mi = min(mi, v);
            return;
        }
        if (k < sz/2) {
            l->set(k, v);
        } else {
            r->set(k-sz/2, v);
        }
        mi = min(l->mi, r->mi);
    }
    int get(int a, int b) {
        if (b <= 0 || sz <= a) return INF;
        if (a <= 0 && sz <= b) return mi;
        return min(l->get(a, b), r->get(a-sz/2, b-sz/2));
    }
};

const int MN = 200010;
int n, q;
int b[MN];
vector<int> v[MN];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> n;
    int ans = 0;
    for (int i = 0; i < n; i++) {
        cin >> b[i];
        if (b[i] == 0) {
            ans++;
            b[i]--;
        }
    }
    cin >> q;
    for (int i = 0; i < q; i++) {
        int l, r;
        cin >> l >> r; l--;
        v[l].push_back(r);
    }
    Node* tr = new Node(n+1);
    for (int i = 0; i <= n; i++) {
        if (i) {
            tr->set(i, tr->get(i-1, i) + b[i-1]);
        } else {
            tr->set(0, 0);
        }
        for (int j = 0; j < int(v[i].size()); j++) {
            int r = v[i][j];
            tr->set(r, tr->get(i, r));
        }
    }
    ans += tr->get(n, n+1);
    cout << ans << endl;
    return 0;
}
### 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 <bits/stdc++.h>

using namespace std;

const int INF = 2e9;

struct Node {
    typedef Node* NP;
    NP l, r;
    int sz, mi;
    Node(int _sz) : l(NULL), r(NULL), sz(_sz) {
        mi = INF;
        if (sz == 1) return;
        l = new Node(sz/2);
        r = new Node(sz-sz/2);
    }
    void set(int k, int v) {
        if (sz == 1) {
            mi = min(mi, v);
            return;
        }
        if (k < sz/2) {
            l->set(k, v);
        } else {
            r->set(k-sz/2, v);
        }
        mi = min(l->mi, r->mi);
    }
    int get(int a, int b) {
        if (b <= 0 || sz <= a) return INF;
        if (a <= 0 && sz <= b) return mi;
        return min(l->get(a, b), r->get(a-sz/2, b-sz/2));
    }
};

const int MN = 200010;
int n, q;
int b[MN];
vector<int> v[MN];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> n;
    int ans = 0;
    for (int i = 0; i < n; i++) {
        cin >> b[i];
        if (b[i] == 0) {
            ans++;
            b[i]--;
        }
    }
    cin >> q;
    for (int i = 0; i < q; i++) {
        int l, r;
        cin >> l >> r; l--;
        v[l].push_back(r);
    }
    Node* tr = new Node(n+1);
    for (int i = 0; i <= n; i++) {
        if (i) {
            tr->set(i, tr->get(i-1, i) + b[i-1]);
        } else {
            tr->set(0, 0);
        }
        for (int j = 0; j < int(v[i].size()); j++) {
            int r = v[i][j];
            tr->set(r, tr->get(i, r));
        }
    }
    ans += tr->get(n, n+1);
    cout << ans << endl;
    return 0;
}
```