You are given a chessboard of size n×nn×n. It is filled with numbers from 11 to n2n2 in the following way: the first ?n22??n22? numbers from 11 to ?n22??n22? are written in the cells with even sum of coordinates from left to right from top to bottom. The rest n2??n22?n2??n22? numbers from ?n22?+1?n22?+1 to n2n2 are written in the cells with odd sum of coordinates from left to right from top to bottom. The operation ?xy??xy? means division xx by yy rounded up.

For example, the left board on the following picture is the chessboard which is given for n=4n=4 and the right board is the chessboard which is given for n=5n=5.

You are given qq queries. The ii-th query is described as a pair xi,yixi,yi. The answer to the ii-th query is the number written in the cell xi,yixi,yi (xixi is the row, yiyi is the column). Rows and columns are numbered from 11 to nn.

Input
The first line contains two integers nn and qq (1≤n≤1091≤n≤109, 1≤q≤1051≤q≤105) — the size of the board and the number of queries.

The next qq lines contain two integers each. The ii-th line contains two integers xi,yixi,yi (1≤xi,yi≤n1≤xi,yi≤n) — description of the ii-th query.

Output
For each query from 11 to qq print the answer to this query. The answer to the ii-th query is the number written in the cell xi,yixi,yi (xixi is the row, yiyi is the column). Rows and columns are numbered from 11 to nn. Queries are numbered from 11 to qq in order of the input.

Examples
Input
4 5
1 1
4 4
4 3
3 2
2 4
Output
1
8
16
13
4
Input
5 4
2 1
4 2
3 3
3 4
Output
16
9
7
20
Note
Answers to the queries from examples are on the board in the picture from the problem statement.
這個(gè)題感覺(jué)上就是找規(guī)律，題目的方法很多，我是根據(jù)n的奇偶性，以及（x+y）的奇偶性來(lái)判斷，推斷出式子來(lái)就可以了。注意用長(zhǎng)整型。
代碼如下：

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #define ll long long using namespace std;ll n; ll x,y; ll t;int main() {while(scanf("%I64d%I64d",&n,&t)!=EOF){while(t--){scanf("%I64d%I64d",&x,&y);if(n%2==0){ll num=(x-1)*(n/2)+((y+1)/2);if((x+y)%2==0)printf("%I64d\n",num);else printf("%I64d\n",num+n*n/2);}else{ll num=((x-1)*n+1+y)/2;if((x+y)%2==0)printf("%I64d\n",num);else printf("%I64d\n",num+n*n/2+1);}}}return 0; }

努力加油a啊，(^o)/~

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