#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000005;
const int mod = 1e9 + 7;
long long n, k, l;
int ori[maxn];
int pos[maxn];
long long dp[maxn][2];
long long ans;
int cmp(int a, int b) { return ori[a] < ori[b]; }
int gap;
int now, pre;
long long low;
void cal(int i, int j) {
  long long cnt = (low - j + 1) % mod;
  if (i > gap && l % n) cnt = (cnt - 1 + mod) % mod;
  ans = (ans + dp[i][now] * cnt % mod) % mod;
}
int main() {
  while (cin >> n >> l >> k) {
    ans = 0;
    gap = (l - 1) % n;
    low = l / n;
    if (l % n) low++;
    now = 0, pre = 1;
    for (int i = 0; i < n; i++) {
      scanf("%d", ori + i);
      dp[i][0] = 1;
      pos[i] = i;
      cal(i, 1);
    }
    sort(pos, pos + n, cmp);
    int p;
    long long sum = 0;
    k = min(k, low);
    for (int j = 2; j <= k; j++) {
      p = 0;
      sum = 0;
      now = now ^ 1, pre = pre ^ 1;
      for (int ii = 0; ii < n; ii++) {
        int i = pos[ii];
        while (p < n && ori[pos[p]] <= ori[i]) {
          sum += dp[pos[p]][pre];
          sum %= mod;
          p++;
        }
        dp[i][now] = sum;
        cal(i, j);
      }
    }
    while (ans < 0) ans += mod;
    cout << ans << endl;
  }
}

### Prompt

Generate a CPP solution to the following problem:

While Duff was resting in the beach, she accidentally found a strange array b0, b1, ..., bl - 1 consisting of l positive integers. This array was strange because it was extremely long, but there was another (maybe shorter) array, a0, ..., an - 1 that b can be build from a with formula: bi = ai mod n where a mod b denoted the remainder of dividing a by b.

<image>

Duff is so curious, she wants to know the number of subsequences of b like bi1, bi2, ..., bix (0 ≤ i1 < i2 < ... < ix < l), such that: 

  * 1 ≤ x ≤ k
  * For each 1 ≤ j ≤ x - 1, <image>
  * For each 1 ≤ j ≤ x - 1, bij ≤ bij + 1. i.e this subsequence is non-decreasing. 



Since this number can be very large, she want to know it modulo 109 + 7.

Duff is not a programmer, and Malek is unavailable at the moment. So she asked for your help. Please tell her this number.

Input

The first line of input contains three integers, n, l and k (1 ≤ n, k, n × k ≤ 106 and 1 ≤ l ≤ 1018).

The second line contains n space separated integers, a0, a1, ..., an - 1 (1 ≤ ai ≤ 109 for each 0 ≤ i ≤ n - 1). 

Output

Print the answer modulo 1 000 000 007 in one line.

Examples

Input

3 5 3
5 9 1


Output

10


Input

5 10 3
1 2 3 4 5


Output

25

Note

In the first sample case, <image>. So all such sequences are: <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image> and <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000005;
const int mod = 1e9 + 7;
long long n, k, l;
int ori[maxn];
int pos[maxn];
long long dp[maxn][2];
long long ans;
int cmp(int a, int b) { return ori[a] < ori[b]; }
int gap;
int now, pre;
long long low;
void cal(int i, int j) {
  long long cnt = (low - j + 1) % mod;
  if (i > gap && l % n) cnt = (cnt - 1 + mod) % mod;
  ans = (ans + dp[i][now] * cnt % mod) % mod;
}
int main() {
  while (cin >> n >> l >> k) {
    ans = 0;
    gap = (l - 1) % n;
    low = l / n;
    if (l % n) low++;
    now = 0, pre = 1;
    for (int i = 0; i < n; i++) {
      scanf("%d", ori + i);
      dp[i][0] = 1;
      pos[i] = i;
      cal(i, 1);
    }
    sort(pos, pos + n, cmp);
    int p;
    long long sum = 0;
    k = min(k, low);
    for (int j = 2; j <= k; j++) {
      p = 0;
      sum = 0;
      now = now ^ 1, pre = pre ^ 1;
      for (int ii = 0; ii < n; ii++) {
        int i = pos[ii];
        while (p < n && ori[pos[p]] <= ori[i]) {
          sum += dp[pos[p]][pre];
          sum %= mod;
          p++;
        }
        dp[i][now] = sum;
        cal(i, j);
      }
    }
    while (ans < 0) ans += mod;
    cout << ans << endl;
  }
}

```