You have an error in your SQL syntax

Here I am faced with such a problem, what is the error? I checked the code more than once

register.php

<div class="content"> <form method="POST" action="/register" class="reg_form"> <br><input type="text" name="login" > - Логин <br><input type="password" name="password" > -Пароль <br><input type="email" name="email" > - E-Mail <br><br><input type="submit" name="register" value="Регистрация"> </form> <?php if (isset($_POST['register'])) { require_once 'assets/reg.php'; } ?> </div> 

reg.php

 $_POST["login"] = FormChars($_POST["login"]); $_POST["email"] = FormChars($_POST["email"]); $_POST["password"] = GenPass(FormChars($_POST["password"]),$_POST['login']); $login = $_POST["login"]; $email = $_POST["email"]; $pass = $_POST["password"]; if (!$login or !$email or !$pass) exit("<div class='alert'><div class='left_block'></div>Не все поля заполнены!</div>"); $row = mysqli_fetch_assoc(mysqli_query($CONNECT, "SELECT username FROM users WHERE username = '$login'")); if ($row['username']) exit("<div class='alert'>Данное имя пользователя уже используется!</div>"); $row = mysqli_fetch_assoc(mysqli_query($CONNECT, "SELECT email FROM users WHERE email = '$email'")); if ($row['email']) exit("<div class='alert'>Данный адрес электронной почты уже зарегистрирован!</div>"); mysqli_query($CONNECT, "INSERT INTO 'users' VALUES ('','$_POST[login]','$_POST[password]','$_POST[email]','NOW()'") or die(mysqli_error($CONNECT)); echo "<div class='success'>Поздравляем, вы успешно зарегистрировались!</div>";