A. Shaass and Oskols

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Shaass has decided to hunt some birds. There are n horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to n from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are a_i oskols sitting on the i-th wire.

Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the i-th wire). Consequently all the birds on the i-th wire to the left of the dead bird get scared and jump up on the wire number i - 1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number i + 1, if there exists no such wire they fly away.

Shaass has shot m birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.

Input

The first line of the input contains an integer n, (1 ≤ n ≤ 100). The next line contains a list of space-separated integersa₁, a₂, ..., a_n, (0 ≤ a_i ≤ 100).

The third line contains an integer m, (0 ≤ m ≤ 100). Each of the next m lines contains two integers x_i and y_i. The integers mean that for the i-th time Shaass shoot the y_i-th (from left) bird on the x_i-th wire, (1 ≤ x_i ≤ n, 1 ≤ y_i). It's guaranteed there will be at least y_i birds on the x_i-th wire at that moment.

Output

On the i-th line of the output print the number of birds on the i-th wire.

Sample test(s)

input

510 10 10 10 1052 53 132 121 134 6

output

0125016

input

32 4 112 2

output

303

解题说明：此题只需要记住状态转移即可

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int main(){int n,i,m,t;int pos,up;int a[101];scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&a[i]);}scanf("%d",&m);for(t=0;t<m;t++){scanf("%d %d",&pos,&up);if(pos>1){a[pos-2]+=up-1;}if(pos<n){a[pos]+=a[pos-1]-up;}a[pos-1]=0;}for(i=0;i<n;i++){printf("%d\n",a[i]);}return 0;}