Problem: There are n cells, numbered 1,2,..., n from left to right. You have to place a robot at any cell initially. The robot must make exactly k moves.

In one move, the robot must move one cell to the left or right, provided that it doesn't move out of bounds. In other words, if the robot was in the cell i, it must move to either the cell i-1 or the cell i+1, as long as it lies between 1 and n (endpoints inclusive). The cells, in the order they are visited (including the cell the robot is placed), together make a good path.

Each cell i has a value a_i associated with it. Let c_0, c_1, ..., c_k be the sequence of cells in a good path in the order they are visited (c_0 is the cell robot is initially placed, c_1 is the cell where the robot is after its first move, and so on; more formally, c_i is the cell that the robot is at after i moves). Then the value of the path is calculated as a_{c_0} + a_{c_1} + ... + a_{c_k}.

Your task is to calculate the sum of values over all possible good paths. Since this number can be very large, output it modulo 10^9 + 7. Two good paths are considered different if the starting cell differs or there exists an integer i ∈ [1, k] such that the current cell of the robot after exactly i moves is different in those paths.

You must process q updates to a and print the updated sum each time. Each update changes the value of exactly one cell. See the input format and the sample input-output for more details.

Input

The first line of the input contains three space-separated integers n, k and q (2 ≤ n ≤ 5000; 1 ≤ k ≤ 5000; 1 ≤ q ≤ 2 ⋅ 10^5).

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).

q lines follow. Each line contains two space-separated integers i and x (1 ≤ i ≤ n; 1 ≤ x ≤ 10^9) indicating that you must change the value of a_i to x.

Output

Print q integers. The i-th integer should be the sum of values over all good paths after the first i updates are performed. Since the answers may be large, print them modulo 10^9 + 7.

Examples

Input


5 1 5
3 5 1 4 2
1 9
2 4
3 6
4 6
5 2


Output


62
58
78
86
86


Input


5 2 5
3 5 1 4 2
1 9
2 4
3 6
4 6
5 2


Output


157
147
207
227
227


Input


4 40 6
92 21 82 46
3 56
1 72
4 28
1 97
2 49
2 88


Output


239185261
666314041
50729936
516818968
766409450
756910476

Note

In the first example, the good paths are (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4).

Initially the values of a are [3, 5, 1, 4, 2]. After the first update, they become [9, 5, 1, 4, 2]. After the second update, they become [9, 4, 1, 4, 2], and so on.

Solution (in CPP):

Solution:
///****In the name of ALLAH, most Gracious, most Compassionate****//
#include<bits/stdc++.h>

#define Input                   freopen("in.txt","r",stdin)
#define Output                  freopen("out.txt","w",stdout)
#define ll                      long long int
#define ull                     unsigned long long int
#define pii                     pair<int,int>
#define pll                     pair<ll,ll>
#define sc                      scanf
#define scin(x)                 sc("%d",&(x))
#define scin2(x,y)              sc("%d %d",&(x),&(y))
#define scln(x)                 sc("%lld",&(x))
#define scln2(x,y)              sc("%lld %lld",&(x),&(y))
#define pf                      printf
#define all(a)                  (a.begin()),(a.end())
#define UNIQUE(X)               (X).erase(unique(all(X)),(X).end())
#define SORT_UNIQUE(c)          (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define ms(a,b)                 memset(a,b,sizeof(a))
#define pb(a)                   push_back(a)
#define mp                      make_pair
#define db                      double
#define EPS                     10E-10
#define ff                      first
#define ss                      second
#define sqr(x)                  (x)*(x)
#define vi                      vector<int>
#define vl                      vector<ll>
#define vii                     vector<vector<int> >
#define vll                     vector<vector<ll> >
#define DBG                     pf("HI\n")
#define MOD                     1000000007
#define CIN                     ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define RUN_CASE(t,T)           for(__typeof(t) t=1;t<=T;t++)
#define CASE(t)                 printf("Case %d: ",t)
#define CASEl(t)                printf("Case %d:\n",t)
#define intlimit                2147483690
#define longlimit               9223372036854775800
#define infinity                (1<<28)
#define gcd(a,b)                __gcd(a,b)
#define lcm(a,b)                (a)*(b)/gcd(a,b)
#define mxx                     123456789
#define PI                      2*acos(0.0)
#define rep(i,a,b)              for(__typeof(i) i=a;i<=b;i++)
#define rev(i,a,b)              for(__typeof(i) i=a;i>=b;i--)
#define coutv(v)                for(auto it:v)cout<<it<<' ';cout<<'\n';
#define cinv(v)                 for(auto &it:v)cin>>it;
#define srt(v)                  sort(v.begin(),v.end());
#define rsrt(v)                 sort(v.rbegin(),v.rend());
#define rvs(v)                  reverse(v.begin(),v.end());

using namespace std;


/** Debugging Tool **/
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args)
{
    cerr << *it << " = " << a << ' ';
    err(++it, args...);
}


ll n,cellfre[5005],ara[5005],dp[5005][5005],fre[5005][5005];
///dp[i][j] means number of j length path start from i
///We can also say it as dp[i][j] means number of j length path ends at i

ll FuN(ll pos,ll kbaki)
{
    if(kbaki == 0){
        return dp[pos][kbaki] = 1;
    }
    if(dp[pos][kbaki] != -1)
        return dp[pos][kbaki];

    ll ret = 0;
    if(pos > 1)
        ret = (ret + FuN(pos-1 , kbaki-1))%MOD;
    if(pos < n)
        ret = (ret + FuN(pos+1 , kbaki-1))%MOD;

    return dp[pos][kbaki] = ret;
}

void Solve(int t)
{
    ll i,j,k,q,idx,x,ans = 0;
    cin>>n>>k>>q;

    for(i=1 ; i<=n ; i++)
        cin>>ara[i];

    ms(dp , -1);
    for(i=1 ; i<=n ; i++){
        for(int kbaki=0 ; kbaki <= k ; kbaki++){
            fre[i][kbaki] = (FuN(i , kbaki)*FuN(i , k-kbaki))%MOD;
            cellfre[i] = (cellfre[i] + fre[i][kbaki])%MOD;
        }
    }

    for(i=1 ; i<=n ; i++){
        ans = (ans + (cellfre[i] * ara[i])%MOD)%MOD;
    }

    for(i=1 ; i<=q ; i++){
        cin>>idx>>x;

        ans = (ans - (cellfre[idx] * ara[idx])%MOD + MOD)%MOD;
        ara[idx] = x;
        ans = (ans + (cellfre[idx] * ara[idx])%MOD)%MOD;

        cout<<ans<<endl;
    }
}

int main()
{
    CIN;
    int t,T;
    T = 1;
//    cin>>T;
    RUN_CASE(t,T)
    {
        Solve(t);
    }
    return 0;
}

/** Stuff u should look for
    * Avoid overflow
    * Corner(Small) case
    * Don't get stuck on one approach
    * Don't forget to clean used memory in each case
**/