#include <bits/stdc++.h>
using namespace std;

const int INF = 1e9;
const int maxn = 2e5 + 5;
int minv[maxn << 2];
#define lson o*2,l,mid
#define rson o*2+1,mid+1,r
 
void pushup(int o){
    minv[o] = min(minv[o*2],minv[o*2+1]);
}
 
void build(int o, int l, int r){
    minv[o] = INF;
    if(l == r)
        return;
    int mid = (l + r) / 2;
    build(lson);
    build(rson);
}
 
void update(int o, int l, int r, int p, int v){
    if(l == r){
        minv[o] = v;
        return;
    }
    int mid = (l + r) / 2;
    if(p <= mid)
        update(lson, p, v);
    if(p > mid)
        update(rson, p, v);
    pushup(o);
}
 
int query(int o, int l, int r, int ql, int qr){
    if(ql <= l && r <= qr){
        return minv[o];
    }
    int mid = (l + r) / 2;
    if(qr <= mid)
        return query(lson, ql, qr);
    if(ql > mid)
        return query(rson, ql, qr);
    return min(query(lson, ql, qr), query(rson, ql, qr));
}

int n, q;
int b[maxn];
vector<int> G[maxn];

int main(){
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n;
    int delta = 0;
    for(int i = 1;i <= n;i++){
        cin >> b[i];
        delta += (b[i] == 0);
    }
    cin >> q;
    for(int i = 1;i <= q;i++){
        int l, r;
        cin >> l >> r;
        G[l].push_back(r);
    }
    build(1, 1, n + 1);
    update(1, 1, n + 1, 1, 0);
    for(int i = 1;i <= n;i++){
        //vector<pair<int, int> > tmp;
        for(auto r : G[i]){
            int val = query(1, 1, n + 1, i, r + 1);
            update(1, 1, n + 1, r + 1, val);
        }
        int cost = (b[i] == 1) - (b[i] == 0);
        int val = query(1, 1, n + 1, i, i);
        int val1 = query(1, 1, n + 1, i + 1, i + 1); 
        if(val + cost < val1)
            update(1, 1, n + 1, i + 1, val + cost);
    }
    int ans = query(1, 1, n + 1, n + 1, n + 1);
    cout << ans + delta << endl;
    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 <bits/stdc++.h>
using namespace std;

const int INF = 1e9;
const int maxn = 2e5 + 5;
int minv[maxn << 2];
#define lson o*2,l,mid
#define rson o*2+1,mid+1,r
 
void pushup(int o){
    minv[o] = min(minv[o*2],minv[o*2+1]);
}
 
void build(int o, int l, int r){
    minv[o] = INF;
    if(l == r)
        return;
    int mid = (l + r) / 2;
    build(lson);
    build(rson);
}
 
void update(int o, int l, int r, int p, int v){
    if(l == r){
        minv[o] = v;
        return;
    }
    int mid = (l + r) / 2;
    if(p <= mid)
        update(lson, p, v);
    if(p > mid)
        update(rson, p, v);
    pushup(o);
}
 
int query(int o, int l, int r, int ql, int qr){
    if(ql <= l && r <= qr){
        return minv[o];
    }
    int mid = (l + r) / 2;
    if(qr <= mid)
        return query(lson, ql, qr);
    if(ql > mid)
        return query(rson, ql, qr);
    return min(query(lson, ql, qr), query(rson, ql, qr));
}

int n, q;
int b[maxn];
vector<int> G[maxn];

int main(){
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n;
    int delta = 0;
    for(int i = 1;i <= n;i++){
        cin >> b[i];
        delta += (b[i] == 0);
    }
    cin >> q;
    for(int i = 1;i <= q;i++){
        int l, r;
        cin >> l >> r;
        G[l].push_back(r);
    }
    build(1, 1, n + 1);
    update(1, 1, n + 1, 1, 0);
    for(int i = 1;i <= n;i++){
        //vector<pair<int, int> > tmp;
        for(auto r : G[i]){
            int val = query(1, 1, n + 1, i, r + 1);
            update(1, 1, n + 1, r + 1, val);
        }
        int cost = (b[i] == 1) - (b[i] == 0);
        int val = query(1, 1, n + 1, i, i);
        int val1 = query(1, 1, n + 1, i + 1, i + 1); 
        if(val + cost < val1)
            update(1, 1, n + 1, i + 1, val + cost);
    }
    int ans = query(1, 1, n + 1, n + 1, n + 1);
    cout << ans + delta << endl;
    return 0;
}


```