#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 25;
int T, n, ans;
int cards[maxn];//cards[i]表示i牌有cards[i]张.
int GetAns() {
    int cnt = 0, c[maxn] = {0}; //cnt用来统计带牌的步数,c[i]表示有c[i]个张数为i的牌.

    for (int i = 0; i <= 13; i++)
        c[cards[i]]++;//统计张数为1,2,3,4的牌的个数.

    while (c[4] && c[2] >= 2)
        c[4] -= 1, c[2] -= 2, cnt++; //四带二(两个对子).

    while (c[4] && c[1] >= 2)
        c[4] -= 1, c[1] -= 2, cnt++; //四带二(两张单牌).

    while (c[4] && c[2])
        c[4] -= 1, c[2] -= 1, cnt++; //四带二(一个对子看做两张单牌).

    while (c[3] && c[2])
        c[3] -= 1, c[2] -= 1, cnt++; //三带一(一个对子).

    while (c[3] && c[1])
        c[3] -= 1, c[1] -= 1, cnt++; //三带一(一张单牌).

    return cnt + c[4] + c[3] + c[2] + c[1]; //输出贪心所需要的步骤(带牌+炸弹+三张+对子+单牌).
}
void DFS(int depth) {
    if (depth > ans)
        return ;//剪枝一————当前步数大于答案.

    int t = GetAns(); //贪心带牌(剪枝二).

    if (depth + t < ans)
        ans = depth + t; //当前步数+贪心带牌的步数比答案更优.

    //接下来考虑出顺子以减少步数.
    for (int i = 2; i <= 13; i++) { //三顺.
        int pos = i;

        while (cards[pos] >= 3)
            pos++;//pos表示能连续的三张的最后一张.

        if (pos - i >= 2) { //如果连续三张的长度大于2(能成顺).
            for (int j = i + 1; j <= pos - 1; j++) {
                for (int k = i; k <= j; k++)
                    cards[k] -= 3; //出三顺,更新牌数.

                DFS(depth + 1); //进一步搜索.

                for (int k = i; k <= j; k++)
                    cards[k] += 3; //搜索结束,复原牌数.
            }
        }
    }

    for (int i = 2; i <= 13; i++) { //双顺.
        int pos = i;

        while (cards[pos] >= 2)
            pos++;//pos表示能连续的两张的最后一张.

        if (pos - i >= 3) { //如果连续两张的长度大于3(能成顺).
            for (int j = i + 2; j <= pos - 1; j++) {
                for (int k = i; k <= j; k++)
                    cards[k] -= 2; //出双顺,更新牌数.

                DFS(depth + 1); //进一步搜索.

                for (int k = i; k <= j; k++)
                    cards[k] += 2; //搜索结束,复原牌数.
            }
        }
    }

    for (int i = 2; i <= 13; i++) { //单顺.
        int pos = i;

        while (cards[pos] >= 1)
            pos++;//pos表示能连续的单张的最后一张.

        if (pos - i >= 5) { //如果连续两张的长度大于5(能成顺).
            for (int j = i + 4; j <= pos - 1; j++) {
                for (int k = i; k <= j; k++)
                    cards[k] -= 1; //出单顺,更新牌数.

                DFS(depth + 1); //进一步搜索.

                for (int k = i; k <= j; k++)
                    cards[k] += 1; //搜索结束,复原牌数.
            }
        }
    }
}
int main() {
    //  freopen("landlords.in","r",stdin);
    //  freopen("landlords.out","w",stdout);
    scanf("%d%d", &T, &n);

    while (T--) {
        //初始化.
        memset(cards, 0, sizeof(cards));
        ans = 0x7fffffff;

        for (int i = 1, x, color; i <= n; i++) {
            scanf("%d%d", &x, &color);

            if (x == 1)
                x = 13; //特殊处理A.
            else if (x)
                x--;

            cards[x]++;//统计牌.
        }

        DFS(0);
        printf("%d\n", ans);
    }

    return 0;
}