#include <bits/stdc++.h>
using namespace std;
struct item {
  long long a, p;
} s[1000005];
long long dp[2][1000005], sum[2][1000005];
long long n, l, k;
bool cmp(const item &x, const item &y) { return x.a < y.a; }
int main() {
  while (scanf("%I64d %I64d %I64d", &n, &l, &k) != EOF) {
    for (int i = 1; i <= n; i++) {
      scanf("%I64d", &s[i].a);
      s[i].p = i;
    }
    if (l <= n) {
      printf("%I64d\n", l);
      continue;
    }
    sort(s + 1, s + 1 + n, cmp);
    long long nT = l / n, fin = (l - 1) % n + 1;
    if (l % n) nT++;
    int cur = 0, last;
    long long ans = l % 1000000007;
    for (int i = 1; i <= n; i++) {
      dp[cur][i] = 1;
    }
    k = min(k, nT);
    for (int x = 2; x <= k; x++) {
      last = cur;
      cur ^= 1;
      long long sum = 0;
      int pos = 1;
      for (int i = 1; i <= n; i++) {
        while (pos <= n && s[pos].a <= s[i].a) {
          sum = (sum + dp[last][pos++]) % 1000000007;
        }
        dp[cur][i] = sum;
        if (s[i].p <= fin) {
          long long var = (nT - x + 1 + 1000000007) % 1000000007;
          ans = (ans + (dp[cur][i] * var) % 1000000007) % 1000000007;
        } else {
          long long var = (nT - x + 1000000007) % 1000000007;
          ans = (ans + (dp[cur][i] * var) % 1000000007) % 1000000007;
        }
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Your task is to create 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;
struct item {
  long long a, p;
} s[1000005];
long long dp[2][1000005], sum[2][1000005];
long long n, l, k;
bool cmp(const item &x, const item &y) { return x.a < y.a; }
int main() {
  while (scanf("%I64d %I64d %I64d", &n, &l, &k) != EOF) {
    for (int i = 1; i <= n; i++) {
      scanf("%I64d", &s[i].a);
      s[i].p = i;
    }
    if (l <= n) {
      printf("%I64d\n", l);
      continue;
    }
    sort(s + 1, s + 1 + n, cmp);
    long long nT = l / n, fin = (l - 1) % n + 1;
    if (l % n) nT++;
    int cur = 0, last;
    long long ans = l % 1000000007;
    for (int i = 1; i <= n; i++) {
      dp[cur][i] = 1;
    }
    k = min(k, nT);
    for (int x = 2; x <= k; x++) {
      last = cur;
      cur ^= 1;
      long long sum = 0;
      int pos = 1;
      for (int i = 1; i <= n; i++) {
        while (pos <= n && s[pos].a <= s[i].a) {
          sum = (sum + dp[last][pos++]) % 1000000007;
        }
        dp[cur][i] = sum;
        if (s[i].p <= fin) {
          long long var = (nT - x + 1 + 1000000007) % 1000000007;
          ans = (ans + (dp[cur][i] * var) % 1000000007) % 1000000007;
        } else {
          long long var = (nT - x + 1000000007) % 1000000007;
          ans = (ans + (dp[cur][i] * var) % 1000000007) % 1000000007;
        }
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```