#include <bits/stdc++.h>
using namespace std;
inline int in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline string get() {
  char ch[1000010];
  scanf("%s", ch);
  return ch;
}
const long long maxn = 1e6 + 10;
const long long MAX_LG = 21;
const long long INF = 1e9;
const long long SQ = 317;
const long long mod = 1e9 + 7;
map<long long, long long> dp[maxn];
long long n, l, k, cnt, rem, a[maxn], b[maxn], res;
inline bool cmp(const long long &f, const long long &s) { return a[f] < a[s]; }
inline void add(long long i, long long j) {
  long long c = (cnt - j + 1) % mod;
  if (c <= 0) return;
  if (i > rem && l % n) {
    if (c <= 1) return;
    c = (c - 1 + mod) % mod;
  }
  res = (res + dp[i][j] * c % mod) % mod;
}
int32_t main() {
  cin >> n >> l >> k;
  cnt = (l + n - 1) / n, rem = ((l - 1) % n);
  for (long long i = 0; i < n; i++) a[i] = in(), b[i] = i, dp[i][1] = 1;
  sort(b, b + n, cmp);
  for (long long j = 2; j <= k; j++) {
    long long sum = 0, cur = 0;
    for (long long i = 0; i < n; i++) {
      long long vv = b[i];
      while (cur < n && a[b[cur]] <= a[vv])
        sum = (sum + dp[b[cur++]][j - 1]) % mod;
      dp[vv][j] = sum;
    }
  }
  for (long long i = 0; i < n; i++)
    for (long long j = 1; j <= k; j++) {
      add(i, j);
    }
  cout << res << "\n";
}

### 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;
inline int in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline string get() {
  char ch[1000010];
  scanf("%s", ch);
  return ch;
}
const long long maxn = 1e6 + 10;
const long long MAX_LG = 21;
const long long INF = 1e9;
const long long SQ = 317;
const long long mod = 1e9 + 7;
map<long long, long long> dp[maxn];
long long n, l, k, cnt, rem, a[maxn], b[maxn], res;
inline bool cmp(const long long &f, const long long &s) { return a[f] < a[s]; }
inline void add(long long i, long long j) {
  long long c = (cnt - j + 1) % mod;
  if (c <= 0) return;
  if (i > rem && l % n) {
    if (c <= 1) return;
    c = (c - 1 + mod) % mod;
  }
  res = (res + dp[i][j] * c % mod) % mod;
}
int32_t main() {
  cin >> n >> l >> k;
  cnt = (l + n - 1) / n, rem = ((l - 1) % n);
  for (long long i = 0; i < n; i++) a[i] = in(), b[i] = i, dp[i][1] = 1;
  sort(b, b + n, cmp);
  for (long long j = 2; j <= k; j++) {
    long long sum = 0, cur = 0;
    for (long long i = 0; i < n; i++) {
      long long vv = b[i];
      while (cur < n && a[b[cur]] <= a[vv])
        sum = (sum + dp[b[cur++]][j - 1]) % mod;
      dp[vv][j] = sum;
    }
  }
  for (long long i = 0; i < n; i++)
    for (long long j = 1; j <= k; j++) {
      add(i, j);
    }
  cout << res << "\n";
}

```