Gym

Given an R×C grid with each cell containing an integer, find the number of subrectangles in this grid that contain only one distinct integer; this means every cell in a subrectangle contains the same integer.

A subrectangle is defined by two cells: the top left cell (r₁, c₁), and the bottom-right cell (r₂, c₂) (1 ≤ r₁ ≤ r₂ ≤ R) (1 ≤ c₁ ≤ c₂ ≤ C), assuming that rows are numbered from top to bottom and columns are numbered from left to right.

Input

The first line of input contains a single integer T, the number of test cases.

The first line of each test case contains two integers R and C (1 ≤ R, C ≤ 1000), the number of rows and the number of columns of the grid, respectively.

Each of the next R lines contains C integers between 1 and 10⁹, representing the values in the row.

Output

For each test case, print the answer on a single line.

Example

Input

1
3 3
3 3 1
3 3 1
2 2 5

Output

16

题意:问由单一数字组成的矩阵的个数。
思路：
dp[i][j]表示以位置i,j为右下角的矩阵个数。
先处理出当前位置向上延伸相同的数字最高有多高。用num[i]记录。
用单调栈处理出在此之前的，第一个小于自己的num[i].
判断中间有没有插入别的数，再进行处理。详见solve函数。

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define debug(a,i) cout<<#a<<"["<<i<<"] = "<<a[i]<<endl;
#define ls (t<<1)
#define rs ((t<<1)|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1024;
const int maxm = 100086;
const int inf = 2.1e9;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);
int n,m;
int mp[maxn][maxn];
int num[maxn];
ll ans;
struct  node{
    int num,pos;
};
stack<node>st;
int pre[maxn];
int pre1[maxn];
ll dp[maxn];
void solve(int t){
    memset(dp,0,sizeof(dp));
    memset(pre,0,sizeof(pre));
    while(!st.empty()){
        st.pop();
    }
    num[0]=-1;
    for(int i=m;i>=0;i--){
        while(!st.empty()&&st.top().num>num[i]){
            pre[st.top().pos]=i;
            st.pop();
        }
        st.push(node{num[i],i});
    }
    int k=1;
    for(int i=1;i<=m;i++){
        if(mp[t][i]!=mp[t][i-1]){
            k=i;
        }
        pre1[i]=k;
    }
    int pree;
    for(int i=1;i<=m;i++){
        if(pre1[i]>pre[i]){
            dp[i]=1ll*num[i]*(i-pre1[i]+1);
            ans+=dp[i];
        }
        else{
            dp[i]=1ll*num[i]*(i-pre[i])+dp[pre[i]];
            ans+=dp[i];
        }
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf("%d",&mp[i][j]);
            }
        }
        ans=0;
        memset(num,0,sizeof(num));
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(mp[i][j]==mp[i-1][j]){
                    num[j]++;
                }
                else{
                    num[j]=1;
                }
            }
            solve(i);
        }
        printf("%lld
",ans);
    }
    return 0;
}