#include <iostream>
using namespace std;
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <cmath>
#include <cstring>
#include <iomanip>

using ll = long long;

const int N = 310;

int n, m;
int a[N];
int b[N];
vector<int> mm;
int ans[N][N];

void solve()
{
	cin >> n >> m;
	bool flag = false;
	for (int i = 1;i <= m;i++)
	{
		cin >> a[i];
		b[i] = a[i];
		if (a[i] > (n + 1) / 2) flag = true;
		else if (a[i] == (n + 1) / 2)
		{
			if (n % 2 == 1 && !mm.empty() && mm.back() == i - 1) flag = true;
			else mm.push_back(i);
		}
	}

	if (flag) // 某一列棋子过多或在奇数行情况下出现了相邻两列棋子都放满的情况
	{
		cout << "No\n";
		return;
	}

	if (n % 2 == 0 || mm.empty()) // 偶数行或无棋子放满的情况
	{
		cout << "Yes\n";
		for (int j = 1;j <= m;j++)
			for (int i = 1;i <= n;i++)
			{
				if (b[j] && (i + j) % 2 == 0)
				{
					ans[i][j] = 1;
					b[j]--;
				}
				else if (b[j] == 0) break;
			}
		for (int i = 1;i <= n;i++)
		{
			for (int j = 1;j <= m;j++)
				cout << ans[i][j];
			cout << '\n';
		}
		return;
	}

	for (auto j : mm) // 将需要放满的列先放上棋子
		for (int i = 1;i <= n;i += 2)
			ans[i][j] = 1;

	for (int p = 1;p < mm.size();p++)
	{
		int pre = mm[p - 1] + 1, suf = mm[p] - 1;
		if ((mm[p] - mm[p - 1]) % 2 == 1) // 中间有偶数列，在中间列中尽量寻找一个可以不用交替的地方1
		{
			bool flag = false; // 判断是否能找到一个不用交替的地方
			for (int i = 2;i <= n;i += 2)
			{
				if (b[pre])
				{
					ans[i][pre] = 1;
					b[pre]--;
				}
				else break;
			}

			for (int j = 1;j <= suf - pre;j++)
			{
				if (!flag) // 暂未找到可不交替的列
				{
					if (j % 2 == 1) // 上一列放置在偶数行
					{
						if (a[pre + j] + a[pre + j - 1] <= n / 2) // 不用交替
						{
							flag = true;
							for (int i = 1;i <= n;i++)
							{
								if ((i + j) % 2 == 1 && ans[i][j - 1] != 1 && b[pre + j])
								{
									ans[i][pre + j] = 1;
									b[pre + j]--;
								}
								else if (b[pre + j] == 0) break;
							}
						}
						else // 需要交替
						{
							for (int i = 1;i <= n;i++)
							{
								if ((i + j) % 2 == 0 && b[pre + j])
								{
									ans[i][pre + j] = 1;
									b[pre + j]--;
								}
								else if (b[pre + j] == 0) break;
							}
						}
					}
					else // 上一列放置在奇数行
					{
						if (a[pre + j] + a[pre + j - 1] <= n / 2 + 1) // 不用交替
						{
							flag = true;
							for (int i = 1;i <= n;i++)
							{
								if ((i + j) % 2 == 1 && ans[i][j - 1] != 1 && b[pre + j])
								{
									ans[i][pre + j] = 1;
									b[pre + j]--;
								}
								else if (b[pre + j] == 0) break;
							}
						}
						else // 需要交替
						{
							for (int i = 1;i <= n;i++)
							{
								if ((i + j) % 2 == 0 && b[pre + j])
								{
									ans[i][pre + j] = 1;
									b[pre + j]--;
								}
								else if (b[pre + j] == 0) break;
							}
						}
					}
				}
				else // 已找到可交替的列
				{
					for (int i = 1;i <= n;i++)
					{
						if ((i + j) % 2 == 1 && b[pre + j])
						{
							ans[i][pre + j] = 1;
							b[pre + j]--;
						}
						else if (b[pre + j] == 0) break;
					}
				}
			}
			if (!flag) // 未找到满足条件的位置，无解
			{
				cout << "No\n";
				return;
			}
		}
		else // 中间有奇数列，直接交替放置
		{
			for (int j = 0;j <= suf - pre;j++)
			{
				for (int i = 1;i <= n;i++)
				{
					if ((i + j) % 2 == 0 && b[pre + j])
					{
						ans[i][pre + j] = 1;
						b[pre + j]--;
					}
					else if (b[pre + j] == 0) break;
				}
			}
		}
	}
	int pre = mm.front() - 1, suf = mm.back() + 1;
	//cout << pre << ' ' << suf << '\n';
	//cout << b[4] << '\n';
	for (int j = 0;pre - j > 0;j++)
	{
		for (int i = 1;i <= n;i++)
			if ((i + j) % 2 == 0 && b[pre - j])
			{
				ans[i][pre - j] = 1;
				b[pre - j]--;
			}
			else if (b[pre - j] == 0) break;
	}
	for (int j = 0;suf + j <= m;j++)
	{
		for (int i = 1;i <= n;i++)
			if ((i + j) % 2 == 0 && b[suf + j])
			{
				ans[i][suf + j] = 1;
				b[suf + j]--;
			}
			else if (b[suf + j] == 0) break;
	}
	cout << "Yes\n";
	for (int i = 1;i <= n;i++)
	{
		for (int j = 1;j <= m;j++)
			cout << ans[i][j];
		cout << '\n';
	}
}

int main()
{
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int T = 1;
	//cin >> T;
	while (T--)
	{
		solve();
	}
	return 0;
}