#include #include #include #include #include #include #include #include #include #include #include #include #include #include #define rep(x, to) for (int x = 0; x < (to); x++) #define REP(x, a, to) for (int x = (a); x < (to); x++) #define foreach(itr, x) for (typeof((x).begin()) itr = (x).begin(); itr != (x).end(); itr++) using namespace std; typedef long long ll; typedef pair PII; typedef pair PLL; int N, T; int a[105]; int dp[105][105]; int dp2[105][105]; int dfs(int i, int j, int tdp[105][105]) { if (tdp[i][j] != 0) return tdp[i][j]; int res = 1; int k_not_exist = 1; for (int k = 0; k < j; k++) { if (a[j] < a[i] && a[k] < a[j] && a[i]-a[j] > a[j]-a[k]) { res = max(res, 1 + dfs(j, k, tdp)); k_not_exist = 0; } } res += k_not_exist; return tdp[i][j] = res; } int main() { cin >> T; while (T--) { int ans = 1; memset(dp, 0, sizeof(dp)); memset(dp2, 0, sizeof(dp2)); cin >> N; rep(i, N) cin >> a[i]; //up for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { if (i == j) dp[i][j] = 1; if (a[i] < a[j]) dp[i][j] = 1; } } for (int j = 0; j < N; j++) { for (int i = 0; i <= j; i++) { if (a[i] >= a[j]) continue; for (int k = 0; k < i; k++) { if (a[i]-a[k] < a[j]-a[i]) { dp[i][j] = max(dp[i][j], dp[k][i]); } } dp[i][j] += 1; } } #if 0 puts("up"); rep(i, 5) rep(j, 5) { printf("%d%c", dp[i][j], j == 5-1 ? '\n' : ' '); } #endif //down reverse(a, a+N); for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { if (i == j) dp2[i][j] = 1; if (a[i] < a[j]) dp2[i][j] = 1; } } for (int j = 0; j < N; j++) { for (int i = 0; i <= j; i++) { if (a[i] >= a[j]) continue; for (int k = 0; k < i; k++) { if (a[i]-a[k] < a[j]-a[i]) { dp2[i][j] = max(dp2[i][j], dp2[k][i]); } } dp2[i][j] += 1; } } #if 0 puts("down"); rep(i, 5) rep(j, 5) { printf("%d%c", dp2[i][j], j == 5-1 ? '\n' : ' '); } #endif for (int j = 0; j < N; j++) { for (int i = 0; i <= j; i++) { for (int k = 0; k <= N-j-1; k++) { if (dp[i][j] + dp2[k][N-j-1] - 1 > ans) { ans = dp[i][j] + dp2[k][N-j-1]-1; //printf("dp[%d][%d]+dp2[%d][%d]-1=%d+%d-1=%d\n",i,j,k,N-j-1,dp[i][j],dp2[k][N-j-1],dp[i][j]+dp2[k][N-j-1]-1); } } } } cout << ans << endl; } return 0; }