//FAFU 1100 线段树 二维线段树 单点更新 区间求和/*题意:一个矩阵，初始化为0，两种操作：1、将某点增加val2、查询一个子矩阵的和思路：二维线段树，单点更新，区间求和，记得pushup.*/#include<stdio.h>#include<string.h>#include<stdlib.h>#define N 1050#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,rint sum[N*3][N*3];int n;void SubBuild(int rt,int l,int r,int t){sum[t][rt] = 0;if(l!=r){int mid = (l + r) >> 1;SubBuild(lson,t);SubBuild(rson,t);}}void Build(int rt,int l,int r){SubBuild(1,1,n,rt);if(l!=r){int mid = (l + r) >> 1;Build(lson);Build(rson);}}void SubUpdate(int rt,int l,int r,int y,int val,int t){if(l == r)sum[t][rt] += val;else{int mid = (l + r) >> 1;if(y <= mid) SubUpdate(lson,y,val,t);else SubUpdate(rson,y,val,t);sum[t][rt] = sum[t][rt<<1] + sum[t][rt<<1|1];}}void Update(int rt,int l,int r,int x,int y,int val){SubUpdate(1,1,n,y,val,rt);if(l!=r){int mid = (l + r) >> 1;if(x <= mid) Update(lson,x,y,val);else Update(rson,x,y,val);}}__int64 SubQuery(int rt,int l,int r,int LY,int RY,int t){if(LY <= l && RY >= r)return sum[t][rt];int mid = (l + r) >> 1;__int64 ans = 0;if(LY <= mid) ans += SubQuery(lson,LY,RY,t);if(RY > mid ) ans += SubQuery(rson,LY,RY,t);return ans;}__int64 Query(int rt,int l,int r,int LX,int RX,int LY,int RY){if(LX <= l && RX >= r){return SubQuery(1,1,n,LY,RY,rt);}int mid = (l + r) >> 1;__int64 ans = 0;if(LX <= mid) ans += Query(lson,LX,RX,LY,RY);if(RX > mid ) ans += Query(rson,LX,RX,LY,RY);return ans;}int main(){int op;int a,b,c,d;while(scanf("%d %d",&op,&n)!=EOF){++n;Build(1,1,n);while(scanf("%d",&op)){if(op == 3)break;if(op == 1){scanf("%d %d %d",&a,&b,&c);++a,++b;Update(1,1,n,a,b,c);}else{scanf("%d %d %d %d",&a,&b,&c,&d);++a,++b,++c,++d;printf("%I64d\n",Query(1,1,n,a,c,b,d));}}}return 0;}
